case study application of parabola

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Study Guides > Boundless Algebra

The parabola, parabolas as conic sections, learning objectives, key takeaways.

• A parabola is formed by the intersection of a plane and a right circular cone.
• All parabolas contain a focus, a directrix, and an axis of symmetry. These vary in exact location depending on the equation used to define the parabola.
• Parabolas are frequently used in physics and engineering for things such as the design of automobile headlight reflectors and the paths of ballistic missiles.
• Parabolas are frequently encountered as graphs of quadratic functions, including the very common equation [latex]y=x^2[/latex].
• ballistic : Relating to projectiles moving under their own momentum, air drag, gravity, and sometimes rocket power
• directrix : A line used to define a curve or surface, especially a line from which any point on a parabola curve has a distance equal to the distance from the focus.
• focus : A point inside the parabolic section defined by forming a right triangle with the axis of symmetry and the cone's horizontal radius.
• axis of symmetry : A line that divides the parabola into two equal halves and also passes through the vertex of the parabola.
• vertex : The point where the plane intersects the exterior surface of the right circular cone, forming one end of the parabola.

Features of Parabolas

Applications of the parabola.

• Projectiles and missiles follow approximately parabolic paths. They are approximate because real-world imperfections affect the movements of objects.
• Parabolic reflectors are common in microwave and satellite dish receiving and transmitting antennas.
• Paraboloids are also observed in the surface of a liquid confined to a container and rotated around the central axis.
• projectile : Any object propelled through space by the application of a force.
• ballistics : The science of objects that predominately fly under the effects of gravity, such as bullets, missiles, or rockets.
• conical : Shaped like a cone; of or relating to a cone or cones.

Real World Parabolic Trajectories

Parabolic Reflectors

Liquid paraboloids, the "vomit comet", licenses & attributions, cc licensed content, shared previously.

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Applications of Parabola in Real-Life

A parabola is a symmetrical curve formed by the points that are equidistant from a fixed point (the focus) and a fixed straight line (the directrix). It’s commonly seen in mathematics and physics and has a U-shape. The path of a thrown object, like a ball, follows a parabolic trajectory under ideal conditions. The parabola has a unique property that makes it very useful in many real-world applications.

In this article, we will learn about, Parabola definitions, real-life applications of parabola, and others in detail.

Parabola Definition

A parabola is a set of all points in a plane that are equidistant from a fixed point (the focus) and a fixed line (the directrix). The vertex of the parabola is the point halfway between the focus and the directrix.

Example: Graph of the equation y = x 2 is a parabola. Image added below shows a parabola:

Parabola Image

Real-Life Applications of Parabola

Parabolas are used in various fields such as physics, engineering, and computer graphics because of their unique properties. Parabolic shape has various real life applications and some of its application are in:

Satellite Dishes

• Solar Panels, etc.

Let’s learn about the same in detail.

Satellite dishes are shaped like parabolas. The receiver is placed at the focus of the parabola so that all signals that hit the dish are reflected to the receiver.

The reflectors in car headlights are parabolic. The light bulb is placed at the focus of the parabola, so the light rays are reflected out in parallel beams.

When a basketball is thrown in a perfect arc, the path it follows is a parabola. This is due to the force of gravity acting on the ball, and the parabola goes perfectly to the basket.

Some bridges are built in the shape of a parabola because of the parabola’s strength and load distribution properties.

We see fountains in our daily life and water in a fountain follows a parabolic path. This is due to the force of gravity acting on the water.

Mirrors in telescopes are often parabolic. Light entering the telescope is reflected by the mirror and focuses at a single point.

Solar Panels

Solar panels often use parabolic mirrors to focus sunlight onto a small area to generate heat.

Uses of Parabola

Below is the uses of Parabola.

• Physics : Parabolic reflectors are used in devices like satellite dishes, telescopes, and microphones to collect or focus waves, such as light, radio, or sound waves, to a single point. This enables efficient transmission or reception of signals.
• Engineering : Parabolic shapes are utilized in the design of bridges, arches, and other structures to evenly distribute weight and withstand stress. They are also employed in designing lenses for optical devices like cameras and projectors.
• Mathematics : Parabolas are fundamental in algebra and calculus, serving as examples for studying quadratic functions, conic sections, and the properties of curves. They provide insights into solving equations and optimizing various mathematical models.
• Astronomy : The orbits of celestial bodies, such as comets and planets, often follow paths that approximate parabolic shapes. Understanding these trajectories is crucial in celestial mechanics for predicting and analyzing the motion of objects in space.

Applications of Parabola – FAQs

What is a parabola in math.

A parabola is a curve produced by the intersection of a right circular cone and a plane parallel to an element of the cone. It is a plane curve defined as the path of a point moving so that its distance from a fixed line (the directrix) is equal to its distance from a fixed point (the focus).

Why are satellite dishes parabolic?

Parabolic reflector antennas, also known as satellite dishes, are used in maritime satellite communications.

How is a parabola used in headlights?

Parabolic reflectors are used in headlights to focus light beams from bulbs, improving driving vision.

Why does a basketball follow a parabolic path?

After you throw it, a basketball follows a motion known as a parabola. The ball begins to move up, but then gravity eventually pulls it back down.

Why are some bridges parabolic?

Parabolas are often found in architecture, especially in the cables of suspension bridges. This is because the stresses on the cables as the bridge is suspended from the top of the towers are most efficiently distributed along a parabola.

How does a fountain create a parabolic path?

In a fountain, the water shot follows a parabolic trajectory as gravity pulls it back down.

Why are mirrors in telescopes parabolic?

Parabolic shape is such that incident parallel rays will converge at a single focal point no matter where on the surface of the mirror they actually strike. This is why the parabolic mirror is the key component of a reflecting telescope along with many other devices designed to focus light.

How do solar panels use parabolas?

Solar Parabolic Dishes work by focusing sunlight onto a central receiver, where it is absorbed and transformed into heat using a parabolic reflector. The parabolic dish has two functions: either collecting or refracting solar energy.

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On this page

• Plane Analytical Geometry
• 1. Distance Formula
• 1a. Gradient (Slope) of a Line, and Inclination
• 1b. Parallel Lines
• 1c. Perpendicular Lines
• 2. The Straight Line
• Perpendicular Distance from a Point to a Line
• 3. The Circle

4. The Parabola

• 4a. Interactive parabola graphs
• 5. The Ellipse
• 5a. Interactive ellipse graphs
• 6. The Hyperbola
• 6a. Interactive hyperbola graphs
• 6b. Conic Sections Summary
• 6c. Conic Sections 3D interactive graph
• 7. Polar Coordinates
• 8. Curves in Polar Coordinates
• Equi-angular Spiral
• Squaring the Circle: rope method with proof

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Why study the parabola?

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Definition of a parabola Formula of a parabola Arch Bridges Horizontal Axis Shifting the Vertex Applications

The parabola has many applications in situations where:

• Radiation often needs to be concentrated at one point (e.g. radio telescopes, pay TV dishes, solar radiation collectors); or
• Radiation needs to be transmitted from a single point into a wide parallel beam (e.g. headlight reflectors).

Here is an animation showing how parallel radio waves are collected by a parabolic antenna. The parallel rays reflect off the antenna and meet at a point (the red dot, labelled F), called the focus .

Click the "See more" button to see more examples. Each time you run it, the dish will become flatter.

Observe that the focus point, F, moves further away from the dish each time you run it.

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Definition of a Parabola

The parabola is defined as the locus of a point which moves so that it is always the same distance from a fixed point (called the focus ) and a given line (called the directrix ).

[The word locus means the set of points satisfying a given condition. See some background in Distance from a Point to a Line .]

In the following graph,

• The focus of the parabola is at `(0, p)`.
• The directrix is the line `y = -p`.
• The focal distance is `|p|` (Distance from the origin to the focus, and from the origin to the directrix. We take absolute value because distance is positive.)
• The point ( x , y ) represents any point on the curve.
• The distance d from any point ( x , y ) to the focus `(0, p)` is the same as the distance from ( x , y ) to the directrix.
• The axis of symmetry of this parabola is the y -axis.

Parabola, showing focus (0, p), and directrix y = − p .

The Formula for a Parabola - Vertical Axis

Adding to our diagram from above, we see that the distance `d = y + p`.

Note that d = y + p .

Now, using the Distance Formula on the general points `(0, p)` and `(x, y)`, and equating it to our value `d = y + p`, we have

`sqrt((x-0)^2+(y-p)^2)=y+p`

Squaring both sides gives:

( x − 0) 2 + ( y − p ) 2 = ( y + p ) 2

Simplifying gives us the formula for a parabola :

In more familiar form, with " y = " on the left, we can write this as:

`y=x^2/(4p)`

where p is the focal distance of the parabola.

Now let's see what "the locus of points equidistant from a point to a line" means.

Each of the colour-coded line segments is the same length in this spider-like graph:

Each colored segment has the same length.

Don't miss Interactive Parabola Graphs , where you can explore concepts like focus, directrix and vertex.

Example 1 - Parabola with Vertical Axis

Need graph paper.

Sketch the parabola

Find the focal length and indicate the focus and the directrix on your graph.

The focal length is found by equating the general expression for y

and our particular example:

So we have:

`x^2/(4p)=x^2/2`

This gives `p = 0.5`.

So the focus will be at `(0, 0.5)` and the directrix is the line `y = -0.5`.

Our curve is as follows:

Parabola y = 0.5 x 2 .

Note: Even though the sides look as though they become straight as x increases, in fact they do not. The sides of a parabola just get steeper and steeper (but are never vertical, either).

Arch Bridges − Almost Parabolic

[Actually, such bridges are normally in the shape of a catenary , but that is beyond the scope of this chapter. See Is the Gateway Arch a Parabola? ]

Parabolas with Horizontal Axis

We can also have the situation where the axis of the parabola is horizontal:

Parabola with horizontal axis.

In this case, we have the relation: (not function)

[In a relation , there are two or more values of y for each value of x . On the other hand, a function only has one value of y for each value of x .]

The above graph's axis of symmetry is the x -axis.

Example 2 - Parabola with Horizontal Axis

Sketch the curve and find the equation of the parabola with focus (−2,0) and directrix x = 2.

In this case, we have the following graph:

Parabola `y^2 = -8`

After sketching, we can see that the equation required is in the following form, since we have a horizontal axis:

`y^2 = 4px`

Since `p = -2` (from the question), we can directly write the equation of the parabola:

`y^2 = -8x`

Shifting the Vertex of a Parabola from the Origin

This is a similar concept to the case when we shifted the centre of a circle from the origin.

To shift the vertex of a parabola from (0, 0) to ( h , k ), each x in the equation becomes ( x − h ) and each y becomes ( y − k ) .

So if the axis of a parabola is vertical , and the vertex is at ( h , k ), we have

( x − h ) 2 = 4 p ( y − k )

Parabola ( x − h ) 2 = 4 p ( y − k )

In the above case, the axis of symmetry is the vertical line through the point (h, k) , that is x = h .

If the axis of a parabola is horizontal , and the vertex is at ( h , k ), the equation becomes

( y − k ) 2 = 4 p ( x − h )

Parabola ( y − k ) 2 = 4 p ( x − h )

In the above case, the axis of symmetry is the horizontal line through the point (h, k) , that is y = k .

1. Sketch `x^2= 14y`

`p = 14/4 = 3.5`

So the focus is at `(0, 3.5)`

Directrix: `y = -3.5`

Parabola y = x 2 /14.

2. Find the equation of the parabola having vertex (0, 0), axis along the x -axis and passing through (2, −1).

The curve must have the following orientation, since we know it has horizontal axis and passes through `(2, -1)`:

Parabola, vertex (0, 0), passing through (2, −1).

So we need to use the general formula for a parabola with horizontal axis:

We need to find `p`. We know the curve goes through `(2, -1)`, so we substitute:

`(-1)^2= 4(p)(2) `

→ `1 = 8p`

→ `p = 1/8`.

So the required equation is `y^2=x/2`.

3. We found above that the equation of the parabola with vertex ( h , k ) and axis parallel to the y -axis is

`(x − h)^2= 4p(y − k)`.

Sketch the parabola for which `(h, k)` is ` (-1,2)` and `p= -3`.

We will have vertex at `(-1,2)` and `p = -3` (so the parabola will be "upside down").

The vertex is at `(-1, 2)`, since we know the focal distance is | p | = 3.

Parabola for which `(h, k)` is ` (-1,2)` and `p= -3`.

We don't really need to find the equation, but as an exercise:

`(x − h)^2 = 4p(y − k) `

`(x + 1)^2 = 4(-3)(y − 2) ` `x^2 + 2x + 1 = -12y + 24` `12y = -x^2 − 2x − 1 + 24` `y = (-x^2 − 2x + 23)/12`

Helpful article and graph interactives

See also: How to draw y ^2 = x − 2 ? , which has an extensive explanation of how to manipulate parabola graphs, depending on the formula given.

Also, don't miss Interactive Parabola Graphs , where you can explore parabolas by moving them around and changing parameters.

Applications of Parabolas

Application 1 - antennas.

A parabolic antenna has a cross-section of width 12 m and depth of 2 m. Where should the receiver be placed for best reception?

Parabolic antenna, width 12 m, height 2m.

The receiver should be placed at the focus of the parabolic dish for best reception, because the incoming signal will be concentrated at the focus.

Parabola on the cartesian plane.

We place the vertex of the parabola at the origin (for convenience) and use the equation of the parabola to get the focal distance ( p ) and hence the required point.

In general, the equation for a parabola with vertical axis is

`x^2 = 4py.`

We can see that the parabola passes through the point `(6, 2)`.

Substituting, we have:

`(6)^2 = 4p(2)`

So `p = 36/8 = 4.5`

So we need to place the receiver 4.5 metres from the vertex, along the axis of symmetry of the parabola.

The equation of the parabola is:

`x^2 = 18y `

`y = x^2 /18`

Parabolic antenna showing focus.

Application 2 - Projectiles

A golf ball is dropped and a regular strobe light illustrates its motion as follows...

We observe that it is a parabola . (Well, very close).

What is the equation of the parabola that the golf ball is tracing out?

First we get a set of data points from observing the height of the ball at various times from the graph (I've used the bottom of each circle as the data point):

Using Scientific Notebook, we can model the motion from the data points.

Using one of the Statistics tools in Scientific Notebook (Fit Curve to Data), we obtain:

`y = -0.51429 + 2.804x - 0.15013x^2`

Here's the graph of the model we just found:

Parabolic golf ball motion.

We can use this to find where the ball will be at any time during the motion. For example, when t = 2.5, the golf ball will have height 5.6 m. Also, we can predict when it will next bounce (at around time `18.5`), by solving for `y = 0`.

Using Excel to Model Curves

You can also use Microsoft Excel to module a parabola. After you plot the points, right-click on one of the points and choose "Add Trendline".

Choose Polynomial, degree 2. In "Options" you can get Excel to display the equation of the parabola on the chart.

Conic section: Parabola

All of the graphs in this chapter are examples of conic sections . This means we can obtain each shape by slicing a cone at different angles.

How can we obtain a parabola from slicing a cone?

We start with a double cone (2 right circular cones placed apex to apex):

If we slice a cone parallel to the slant edge of the cone, the resulting shape is a parabola, as shown.

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Analytic Geometry

The parabola, learning objectives.

In this section, you will:

• Graph parabolas with vertices at the origin.
• Write equations of parabolas in standard form.
• Graph parabolas with vertices not at the origin.
• Solve applied problems involving parabolas.

Figure 1. The Olympic torch concludes its journey around the world when it is used to light the Olympic cauldron during the opening ceremony. (credit: Ken Hackman, U.S. Air Force)

Did you know that the Olympic torch is lit several months before the start of the games? The ceremonial method for lighting the flame is the same as in ancient times. The ceremony takes place at the Temple of Hera in Olympia, Greece, and is rooted in Greek mythology, paying tribute to Prometheus, who stole fire from Zeus to give to all humans. One of eleven acting priestesses places the torch at the focus of a parabolic mirror (see (Figure) ), which focuses light rays from the sun to ignite the flame.

Parabolic mirrors (or reflectors) are able to capture energy and focus it to a single point. The advantages of this property are evidenced by the vast list of parabolic objects we use every day: satellite dishes, suspension bridges, telescopes, microphones, spotlights, and car headlights, to name a few. Parabolic reflectors are also used in alternative energy devices, such as solar cookers and water heaters, because they are inexpensive to manufacture and need little maintenance. In this section we will explore the parabola and its uses, including low-cost, energy-efficient solar designs.

Graphing Parabolas with Vertices at the Origin

In The Ellipse , we saw that an ellipse is formed when a plane cuts through a right circular cone. If the plane is parallel to the edge of the cone, an unbounded curve is formed. This curve is a parabola. See (Figure) .

Figure 2. Parabola

Like the ellipse and hyperbola , the parabola can also be defined by a set of points in the coordinate plane. A parabola is the set of all points[latex]\,\left(x,y\right)[/latex] in a plane that are the same distance from a fixed line, called the directrix, and a fixed point (the focus) not on the directrix.

In Quadratic Functions , we learned about a parabola’s vertex and axis of symmetry. Now we extend the discussion to include other key features of the parabola. See (Figure) . Notice that the axis of symmetry passes through the focus and vertex and is perpendicular to the directrix. The vertex is the midpoint between the directrix and the focus.

The line segment that passes through the focus and is parallel to the directrix is called the latus rectum. The endpoints of the latus rectum lie on the curve. By definition, the distance[latex]\,d\,[/latex]from the focus to any point[latex]\,P\,[/latex]on the parabola is equal to the distance from[latex]\,P\,[/latex]to the directrix.

Figure 3. Key features of the parabola

To work with parabolas in the coordinate plane , we consider two cases: those with a vertex at the origin and those with a vertex at a point other than the origin. We begin with the former.

Let[latex]\,\left(x,y\right)\,[/latex]be a point on the parabola with vertex[latex]\,\left(0,0\right),[/latex] focus[latex]\,\left(0,p\right),[/latex]and directrix[latex]\,y= -p\,[/latex] as shown in (Figure) . The distance[latex]\,d\,[/latex]from point[latex]\,\left(x,y\right)\,[/latex]to point[latex]\,\left(x,-p\right)\,[/latex] on the directrix is the difference of the y -values:[latex]\,d=y+p.\,[/latex]The distance from the focus[latex]\,\left(0,p\right)\,[/latex]to the point[latex]\,\left(x,y\right)\,[/latex]is also equal to[latex]\,d\,[/latex]and can be expressed using the distance formula .

Set the two expressions for[latex]\,d\,[/latex]equal to each other and solve for[latex]\,y\,[/latex]to derive the equation of the parabola. We do this because the distance from[latex]\,\left(x,y\right)\,[/latex]to[latex]\,\left(0,p\right)\,[/latex]equals the distance from[latex]\,\left(x,y\right)\,[/latex]to[latex]\,\left(x, -p\right).[/latex]

We then square both sides of the equation, expand the squared terms, and simplify by combining like terms.

The equations of parabolas with vertex[latex]\,\left(0,0\right)\,[/latex]are[latex]\,{y}^{2}=4px\,[/latex]when the x -axis is the axis of symmetry and[latex]\,{x}^{2}=4py\,[/latex]when the y -axis is the axis of symmetry. These standard forms are given below, along with their general graphs and key features.

Standard Forms of Parabolas with Vertex (0, 0)

(Figure) and (Figure) summarize the standard features of parabolas with a vertex at the origin.

Figure 5. (a) When[latex]\,p>0\,[/latex]and the axis of symmetry is the x-axis, the parabola opens right. (b) When[latex]\,p<0\,[/latex]and the axis of symmetry is the x-axis, the parabola opens left. (c) When[latex]\,p<0\,[/latex]and the axis of symmetry is the y-axis, the parabola opens up. (d) When[latex]\text{ }p<0\text{ }[/latex]and the axis of symmetry is the y-axis, the parabola opens down.

The key features of a parabola are its vertex, axis of symmetry, focus, directrix, and latus rectum. See (Figure) . When given a standard equation for a parabola centered at the origin, we can easily identify the key features to graph the parabola.

A line is said to be tangent to a curve if it intersects the curve at exactly one point. If we sketch lines tangent to the parabola at the endpoints of the latus rectum, these lines intersect on the axis of symmetry, as shown in (Figure) .

Given a standard form equation for a parabola centered at (0, 0), sketch the graph.

• Determine which of the standard forms applies to the given equation:[latex]\,{y}^{2}=4px\,[/latex]or[latex]\,{x}^{2}=4py.[/latex]
• the axis of symmetry is the x -axis,[latex]\,y=0[/latex]
• set[latex]\,4p\,[/latex]equal to the coefficient of x in the given equation to solve for[latex]\,p.\,[/latex]If[latex]\,p>0,[/latex] the parabola opens right. If[latex]\,p<0,[/latex] the parabola opens left.
• use[latex]\,p\,[/latex]to find the coordinates of the focus,[latex]\,\left(p,0\right)[/latex]
• use[latex]\,p\,[/latex]to find the equation of the directrix,[latex]\,x=-p[/latex]
• use[latex]\,p\,[/latex]to find the endpoints of the latus rectum,[latex]\,\left(p,±2p\right).\,[/latex]Alternately, substitute[latex]\,x=p\,[/latex]into the original equation.
• the axis of symmetry is the y -axis,[latex]\,x=0[/latex]
• set[latex]\,4p\,[/latex]equal to the coefficient of y in the given equation to solve for[latex]\,p.\,[/latex]If[latex]\,p>0,[/latex] the parabola opens up. If[latex]\,p<0,[/latex] the parabola opens down.
• use[latex]\,p\,[/latex]to find the coordinates of the focus,[latex]\,\left(0,p\right)[/latex]
• use[latex]\,p\,[/latex]to find equation of the directrix,[latex]\,y=-p[/latex]
• use[latex]\,p\,[/latex]to find the endpoints of the latus rectum,[latex]\,\left(±2p,p\right)[/latex]
• Plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.

Graphing a Parabola with Vertex (0, 0) and the x -axis as the Axis of Symmetry

Graph[latex]\,{y}^{2}=24x.\,[/latex]Identify and label the focus , directrix , and endpoints of the latus rectum .

The standard form that applies to the given equation is[latex]\,{y}^{2}=4px.\,[/latex]Thus, the axis of symmetry is the x -axis. It follows that:

• [latex]24=4p,[/latex] so[latex]\,p=6.\,[/latex]Since[latex]\,p>0,[/latex] the parabola opens right
• the coordinates of the focus are[latex]\,\left(p,0\right)=\left(6,0\right)[/latex]
• the equation of the directrix is[latex]\,x=-p=-6[/latex]
• the endpoints of the latus rectum have the same x -coordinate at the focus. To find the endpoints, substitute[latex]\,x=6\,[/latex]into the original equation:[latex]\,\left(6,±12\right)[/latex]

Graph[latex]\,{y}^{2}=-16x.\,[/latex]Identify and label the focus, directrix, and endpoints of the latus rectum.

Focus:[latex]\,\left(-4,0\right);\,[/latex]Directrix:[latex]\,x=4;\,[/latex]Endpoints of the latus rectum:[latex]\,\left(-4,±8\right)[/latex]

Graphing a Parabola with Vertex (0, 0) and the y -axis as the Axis of Symmetry

Graph[latex]\,{x}^{2}=-6y.\,[/latex]Identify and label the focus , directrix , and endpoints of the latus rectum .

The standard form that applies to the given equation is[latex]\,{x}^{2}=4py.\,[/latex]Thus, the axis of symmetry is the y -axis. It follows that:

• [latex]-6=4p,[/latex]so[latex]\,p=-\frac{3}{2}.\,[/latex]Since[latex]\,p<0,[/latex] the parabola opens down.
• the coordinates of the focus are[latex]\,\left(0,p\right)=\left(0,-\frac{3}{2}\right)[/latex]
• the equation of the directrix is[latex]\,y=-p=\frac{3}{2}[/latex]
• the endpoints of the latus rectum can be found by substituting[latex]\text{ }y=\frac{3}{2}\text{ }[/latex]into the original equation,[latex]\,\left(±3,-\frac{3}{2}\right)[/latex]

Next we plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola .

Graph[latex]\,{x}^{2}=8y.\,[/latex]Identify and label the focus, directrix, and endpoints of the latus rectum.

Focus:[latex]\,\left(0,2\right);\,[/latex]Directrix:[latex]\,y=-2;\,[/latex]Endpoints of the latus rectum:[latex]\,\left(±4,2\right).[/latex]

Writing Equations of Parabolas in Standard Form

In the previous examples, we used the standard form equation of a parabola to calculate the locations of its key features. We can also use the calculations in reverse to write an equation for a parabola when given its key features.

Given its focus and directrix, write the equation for a parabola in standard form.

• If the given coordinates of the focus have the form[latex]\,\left(p,0\right),[/latex] then the axis of symmetry is the x -axis. Use the standard form[latex]\,{y}^{2}=4px.[/latex]
• If the given coordinates of the focus have the form[latex]\,\left(0,p\right),[/latex] then the axis of symmetry is the y -axis. Use the standard form[latex]\,{x}^{2}=4py.[/latex]
• Multiply[latex]\,4p.[/latex]
• Substitute the value from Step 2 into the equation determined in Step 1.

Writing the Equation of a Parabola in Standard Form Given its Focus and Directrix

What is the equation for the parabola with focus [latex]\,\left(-\frac{1}{2},0\right)\,[/latex]and directrix [latex]\,x=\frac{1}{2}?[/latex]

The focus has the form[latex]\,\left(p,0\right),[/latex] so the equation will have the form[latex]\,{y}^{2}=4px.[/latex]

• Multiplying[latex]\,4p,[/latex] we have[latex]\,4p=4\left(-\frac{1}{2}\right)=-2.[/latex]
• Substituting for[latex]\,4p,[/latex] we have[latex]\,{y}^{2}=4px=-2x.[/latex]

Therefore, the equation for the parabola is[latex]\,{y}^{2}=-2x.[/latex]

What is the equation for the parabola with focus[latex]\,\left(0,\frac{7}{2}\right)\,[/latex]and directrix[latex]\,y=-\frac{7}{2}?[/latex]

[latex]{x}^{2}=14y.[/latex]

Graphing Parabolas with Vertices Not at the Origin

Like other graphs we’ve worked with, the graph of a parabola can be translated. If a parabola is translated[latex]\,h\,[/latex]units horizontally and[latex]\,k\,[/latex]units vertically, the vertex will be[latex]\,\left(h,k\right).\,[/latex]This translation results in the standard form of the equation we saw previously with[latex]\,x\,[/latex]replaced by[latex]\,\left(x-h\right)\,[/latex]and[latex]\,y\,[/latex]replaced by[latex]\,\left(y-k\right).[/latex]

To graph parabolas with a vertex[latex]\,\left(h,k\right)\,[/latex]other than the origin, we use the standard form[latex]\,{\left(y-k\right)}^{2}=4p\left(x-h\right)\,[/latex]for parabolas that have an axis of symmetry parallel to the x -axis, and[latex]\,{\left(x-h\right)}^{2}=4p\left(y-k\right)\,[/latex]for parabolas that have an axis of symmetry parallel to the y -axis. These standard forms are given below, along with their general graphs and key features.

Standard Forms of Parabolas with Vertex ( h , k )

(Figure) and (Figure) summarize the standard features of parabolas with a vertex at a point[latex]\,\left(h,k\right).[/latex]

Figure 9. (a) When[latex]\,p>0,[/latex] the parabola opens right. (b) When[latex]\,p<0,[/latex] the parabola opens left. (c) When[latex]\,p>0,[/latex]the parabola opens up. (d) When[latex]\,p<0,[/latex] the parabola opens down.

Given a standard form equation for a parabola centered at ( h , k ), sketch the graph.

• Determine which of the standard forms applies to the given equation:[latex]\,{\left(y-k\right)}^{2}=4p\left(x-h\right)\,[/latex]or[latex]\,{\left(x-h\right)}^{2}=4p\left(y-k\right).[/latex]
• use the given equation to identify[latex]\,h\,[/latex]and[latex]\,k\,[/latex]for the vertex,[latex]\,\left(h,k\right)[/latex]
• use the value of[latex]\,k\,[/latex]to determine the axis of symmetry,[latex]\,y=k[/latex]
• set[latex]\,4p\,[/latex]equal to the coefficient of[latex]\,\left(x-h\right)\,[/latex]in the given equation to solve for[latex]\,p.\,[/latex]If[latex]\,p>0,[/latex]the parabola opens right. If[latex]\,p<0,[/latex] the parabola opens left.
• use[latex]\,h,k,[/latex] and[latex]\,p\,[/latex]to find the coordinates of the focus,[latex]\,\left(h+p,\text{ }k\right)[/latex]
• use[latex]\,h\,[/latex] and[latex]\,p\,[/latex]to find the equation of the directrix,[latex]\,x=h-p[/latex]
• use[latex]\,h,k,[/latex] and[latex]\,p\,[/latex]to find the endpoints of the latus rectum,[latex]\,\left(h+p,k±2p\right)[/latex]
• use the value of[latex]\,h\,[/latex]to determine the axis of symmetry,[latex]\,x=h[/latex]
• set[latex]\,4p\,[/latex]equal to the coefficient of[latex]\,\left(y-k\right)\,[/latex]in the given equation to solve for[latex]\,p.\,[/latex]If[latex]\,p>0,[/latex] the parabola opens up. If[latex]\,p<0,[/latex] the parabola opens down.
• use[latex]\,h,k,[/latex] and[latex]\,p\,[/latex]to find the coordinates of the focus,[latex]\,\left(h,\text{ }k+p\right)[/latex]
• use[latex]\,k\,[/latex]and[latex]\,p\,[/latex]to find the equation of the directrix,[latex]\,y=k-p[/latex]
• use[latex]\,h,k,[/latex] and[latex]\,p\,[/latex]to find the endpoints of the latus rectum,[latex]\,\left(h±2p,\text{ }k+p\right)[/latex]
• Plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.

Graphing a Parabola with Vertex ( h , k ) and Axis of Symmetry Parallel to the x -axis

Graph[latex]\,{\left(y-1\right)}^{2}=-16\left(x+3\right).\,[/latex]Identify and label the vertex , axis of symmetry , focus , directrix , and endpoints of the latus rectum .

The standard form that applies to the given equation is[latex]\,{\left(y-k\right)}^{2}=4p\left(x-h\right).\,[/latex]Thus, the axis of symmetry is parallel to the x -axis. It follows that:

• the vertex is[latex]\,\left(h,k\right)=\left(-3,1\right)[/latex]
• the axis of symmetry is[latex]\,y=k=1[/latex]
• [latex]-16=4p,[/latex]so[latex]\,p=-4.\,[/latex]Since[latex]\,p<0,[/latex] the parabola opens left.
• the coordinates of the focus are[latex]\,\left(h+p,k\right)=\left(-3+\left(-4\right),1\right)=\left(-7,1\right)[/latex]
• the equation of the directrix is[latex]\,x=h-p=-3-\left(-4\right)=1[/latex]
• the endpoints of the latus rectum are[latex]\,\left(h+p,k±2p\right)=\left(-3+\left(-4\right),1±2\left(-4\right)\right),[/latex] or[latex]\,\left(-7,-7\right)\,[/latex]and[latex]\,\left(-7,9\right)[/latex]

Next we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. See (Figure) .

Graph[latex]\,{\left(y+1\right)}^{2}=4\left(x-8\right).\,[/latex]Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.

Graphing a Parabola from an Equation Given in General Form

Graph[latex]\,{x}^{2}-8x-28y-208=0.\,[/latex]Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.

Start by writing the equation of the parabola in standard form. The standard form that applies to the given equation is[latex]\,{\left(x-h\right)}^{2}=4p\left(y-k\right).\,[/latex]Thus, the axis of symmetry is parallel to the y -axis. To express the equation of the parabola in this form, we begin by isolating the terms that contain the variable[latex]\,x\,[/latex]in order to complete the square.

It follows that:

• the vertex is[latex]\,\left(h,k\right)=\left(4,-8\right)[/latex]
• the axis of symmetry is[latex]\,x=h=4[/latex]
• since[latex]\,p=7,p>0\,[/latex]and so the parabola opens up
• the coordinates of the focus are[latex]\,\left(h,k+p\right)=\left(4,-8+7\right)=\left(4,-1\right)[/latex]
• the equation of the directrix is[latex]\,y=k-p=-8-7=-15[/latex]
• the endpoints of the latus rectum are[latex]\,\left(h±2p,k+p\right)=\left(4±2\left(7\right),-8+7\right),[/latex] or[latex]\,\left(-10,-1\right)\,[/latex]and[latex]\,\left(18,-1\right)[/latex]

Graph[latex]\,{\left(x+2\right)}^{2}=-20\left(y-3\right).\,[/latex]Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.

Solving Applied Problems Involving Parabolas

As we mentioned at the beginning of the section, parabolas are used to design many objects we use every day, such as telescopes, suspension bridges, microphones, and radar equipment. Parabolic mirrors, such as the one used to light the Olympic torch, have a very unique reflecting property. When rays of light parallel to the parabola’s axis of symmetry are directed toward any surface of the mirror, the light is reflected directly to the focus. See (Figure) . This is why the Olympic torch is ignited when it is held at the focus of the parabolic mirror.

Figure 12. Reflecting property of parabolas

Parabolic mirrors have the ability to focus the sun’s energy to a single point, raising the temperature hundreds of degrees in a matter of seconds. Thus, parabolic mirrors are featured in many low-cost, energy efficient solar products, such as solar cookers, solar heaters, and even travel-sized fire starters.

A cross-section of a design for a travel-sized solar fire starter is shown in (Figure) . The sun’s rays reflect off the parabolic mirror toward an object attached to the igniter. Because the igniter is located at the focus of the parabola, the reflected rays cause the object to burn in just seconds.

• Find the equation of the parabola that models the fire starter. Assume that the vertex of the parabolic mirror is the origin of the coordinate plane.
• Use the equation found in part (a) to find the depth of the fire starter.

Figure 13. Cross-section of a travel-sized solar fire starter

• The vertex of the dish is the origin of the coordinate plane, so the parabola will take the standard form[latex]\,{x}^{2}=4py,[/latex]where[latex]\,p>0.\,[/latex]The igniter, which is the focus, is 1.7 inches above the vertex of the dish. Thus we have[latex]\,p=1.7.\,[/latex] [latex]\begin{array}{ll}{x}^{2}=4py\hfill & \begin{array}{cccc}& & & \end{array}\text{Standard form of upward-facing parabola with vertex (0,0)}\hfill \\ {x}^{2}=4\left(1.7\right)y\hfill & \begin{array}{cccc}& & & \end{array}\text{Substitute 1}\text{.7 for }p.\hfill \\ {x}^{2}=6.8y\hfill & \begin{array}{cccc}& & & \end{array}\text{Multiply}.\hfill \end{array}[/latex]

Balcony-sized solar cookers have been designed for families living in India. The top of a dish has a diameter of 1600 mm. The sun’s rays reflect off the parabolic mirror toward the “cooker,” which is placed 320 mm from the base.

• Find an equation that models a cross-section of the solar cooker. Assume that the vertex of the parabolic mirror is the origin of the coordinate plane, and that the parabola opens to the right (i.e., has the x -axis as its axis of symmetry).
• Use the equation found in part (a) to find the depth of the cooker.
• [latex]{y}^{2}=1280x[/latex]
• The depth of the cooker is 500 mm

Access these online resources for additional instruction and practice with parabolas.

• Conic Sections: The Parabola Part 1 of 2
• Conic Sections: The Parabola Part 2 of 2
• Parabola with Vertical Axis
• Parabola with Horizontal Axis

Key Equations

Key Concepts

• A parabola is the set of all points[latex]\,\left(x,y\right)\,[/latex]in a plane that are the same distance from a fixed line, called the directrix, and a fixed point (the focus) not on the directrix.
• The standard form of a parabola with vertex[latex]\,\left(0,0\right)\,[/latex]and the x -axis as its axis of symmetry can be used to graph the parabola. If[latex]\,p>0,[/latex] the parabola opens right. If[latex]\,p<0,[/latex] the parabola opens left. See (Figure) .
• The standard form of a parabola with vertex[latex]\,\left(0,0\right)\,[/latex]and the y -axis as its axis of symmetry can be used to graph the parabola. If[latex]\,p>0,[/latex] the parabola opens up. If[latex]\,p<0,[/latex] the parabola opens down. See (Figure) .
• When given the focus and directrix of a parabola, we can write its equation in standard form. See (Figure) .
• The standard form of a parabola with vertex[latex]\,\left(h,k\right)\,[/latex]and axis of symmetry parallel to the x -axis can be used to graph the parabola. If[latex]\,p>0,[/latex] the parabola opens right. If[latex]\,p<0,[/latex] the parabola opens left. See (Figure) .
• The standard form of a parabola with vertex[latex]\,\left(h,k\right)\,[/latex]and axis of symmetry parallel to the y -axis can be used to graph the parabola. If[latex]\,p>0,[/latex] the parabola opens up. If[latex]\,p<0,[/latex] the parabola opens down. See (Figure) .
• Real-world situations can be modeled using the standard equations of parabolas. For instance, given the diameter and focus of a cross-section of a parabolic reflector, we can find an equation that models its sides. See (Figure) .

Section Exercises

Define a parabola in terms of its focus and directrix.

A parabola is the set of points in the plane that lie equidistant from a fixed point, the focus, and a fixed line, the directrix.

If the equation of a parabola is written in standard form and[latex]\,p\,[/latex]is positive and the directrix is a vertical line, then what can we conclude about its graph?

If the equation of a parabola is written in standard form and[latex]\,p\,[/latex]is negative and the directrix is a horizontal line, then what can we conclude about its graph?

The graph will open down.

What is the effect on the graph of a parabola if its equation in standard form has increasing values of [latex]\,p\text{?}[/latex]

As the graph of a parabola becomes wider, what will happen to the distance between the focus and directrix?

The distance between the focus and directrix will increase.

For the following exercises, determine whether the given equation is a parabola. If so, rewrite the equation in standard form.

[latex]{y}^{2}=4-{x}^{2}[/latex]

[latex]y=4{x}^{2}[/latex]

[latex]3{x}^{2}-6{y}^{2}=12[/latex]

[latex]{\left(y-3\right)}^{2}=8\left(x-2\right)[/latex]

[latex]{y}^{2}+12x-6y-51=0[/latex]

For the following exercises, rewrite the given equation in standard form, and then determine the vertex[latex]\,\left(V\right),[/latex] focus[latex]\,\left(F\right),[/latex] and directrix[latex]\text{ }\left(d\right)\text{ }[/latex]of the parabola.

[latex]x=8{y}^{2}[/latex]

[latex]{y}^{2}=\frac{1}{8}x,V:\left(0,0\right);F:\left(\frac{1}{32},0\right);d:x=-\frac{1}{32}[/latex]

[latex]y=\frac{1}{4}{x}^{2}[/latex]

[latex]y=-4{x}^{2}[/latex]

[latex]{x}^{2}=-\frac{1}{4}y,V:\left(0,0\right);F:\left(0,-\frac{1}{16}\right);d:y=\frac{1}{16}[/latex]

[latex]x=\frac{1}{8}{y}^{2}[/latex]

[latex]x=36{y}^{2}[/latex]

[latex]{y}^{2}=\frac{1}{36}x,V:\left(0,0\right);F:\left(\frac{1}{144},0\right);d:x=-\frac{1}{144}[/latex]

[latex]x=\frac{1}{36}{y}^{2}[/latex]

[latex]{\left(x-1\right)}^{2}=4\left(y-1\right)[/latex]

[latex]{\left(x-1\right)}^{2}=4\left(y-1\right),V:\left(1,1\right);F:\left(1,2\right);d:y=0[/latex]

[latex]{\left(y-2\right)}^{2}=\frac{4}{5}\left(x+4\right)[/latex]

[latex]{\left(y-4\right)}^{2}=2\left(x+3\right)[/latex]

[latex]{\left(y-4\right)}^{2}=2\left(x+3\right),V:\left(-3,4\right);F:\left(-\frac{5}{2},4\right);d:x=-\frac{7}{2}[/latex]

[latex]{\left(x+1\right)}^{2}=2\left(y+4\right)[/latex]

[latex]{\left(x+4\right)}^{2}=24\left(y+1\right)[/latex]

[latex]{\left(x+4\right)}^{2}=24\left(y+1\right),V:\left(-4,-1\right);F:\left(-4,5\right);d:y=-7[/latex]

[latex]{\left(y+4\right)}^{2}=16\left(x+4\right)[/latex]

[latex]{y}^{2}+12x-6y+21=0[/latex]

[latex]{\left(y-3\right)}^{2}=-12\left(x+1\right),V:\left(-1,3\right);F:\left(-4,3\right);d:x=2[/latex]

[latex]{x}^{2}-4x-24y+28=0[/latex]

[latex]5{x}^{2}-50x-4y+113=0[/latex]

[latex]{\left(x-5\right)}^{2}=\frac{4}{5}\left(y+3\right),V:\left(5,-3\right);F:\left(5,-\frac{14}{5}\right);d:y=-\frac{16}{5}[/latex]

[latex]{y}^{2}-24x+4y-68=0[/latex]

[latex]{x}^{2}-4x+2y-6=0[/latex]

[latex]{\left(x-2\right)}^{2}=-2\left(y-5\right),V:\left(2,5\right);F:\left(2,\frac{9}{2}\right);d:y=\frac{11}{2}[/latex]

[latex]{y}^{2}-6y+12x-3=0[/latex]

[latex]3{y}^{2}-4x-6y+23=0[/latex]

[latex]{\left(y-1\right)}^{2}=\frac{4}{3}\left(x-5\right),V:\left(5,1\right);F:\left(\frac{16}{3},1\right);d:x=\frac{14}{3}[/latex]

[latex]{x}^{2}+4x+8y-4=0[/latex]

For the following exercises, graph the parabola, labeling the focus and the directrix.

[latex]y=36{x}^{2}[/latex]

[latex]y=\frac{1}{36}{x}^{2}[/latex]

[latex]y=-9{x}^{2}[/latex]

[latex]{\left(y-2\right)}^{2}=-\frac{4}{3}\left(x+2\right)[/latex]

[latex]-5{\left(x+5\right)}^{2}=4\left(y+5\right)[/latex]

[latex]-6{\left(y+5\right)}^{2}=4\left(x-4\right)[/latex]

[latex]{y}^{2}-6y-8x+1=0[/latex]

[latex]{x}^{2}+8x+4y+20=0[/latex]

[latex]3{x}^{2}+30x-4y+95=0[/latex]

[latex]{y}^{2}-8x+10y+9=0[/latex]

[latex]{x}^{2}+4x+2y+2=0[/latex]

[latex]{y}^{2}+2y-12x+61=0[/latex]

[latex]-2{x}^{2}+8x-4y-24=0[/latex]

For the following exercises, find the equation of the parabola given information about its graph.

Vertex is[latex]\,\left(0,0\right);[/latex]directrix is[latex]\,y=4,[/latex] focus is[latex]\,\left(0,-4\right).[/latex]

[latex]{x}^{2}=-16y[/latex]

Vertex is[latex]\,\left(0,0\right);\,[/latex]directrix is[latex]\,x=4,[/latex] focus is[latex]\,\left(-4,0\right).[/latex]

Vertex is[latex]\,\left(2,2\right);\,[/latex] directrix is[latex]\,x=2-\sqrt{2},[/latex] focus is[latex]\,\left(2+\sqrt{2},2\right).[/latex]

[latex]{\left(y-2\right)}^{2}=4\sqrt{2}\left(x-2\right)[/latex]

Vertex is[latex]\,\left(-2,3\right);\,[/latex]directrix is[latex]\,x=-\frac{7}{2},[/latex] focus is[latex]\,\left(-\frac{1}{2},3\right).[/latex]

Vertex is[latex]\,\left(\sqrt{2},-\sqrt{3}\right);[/latex] directrix is[latex]\,x=2\sqrt{2},[/latex] focus is[latex]\,\left(0,-\sqrt{3}\right).[/latex]

[latex]{\left(y+\sqrt{3}\right)}^{2}=-4\sqrt{2}\left(x-\sqrt{2}\right)[/latex]

Vertex is[latex]\,\left(1,2\right);\,[/latex]directrix is[latex]\,y=\frac{11}{3},[/latex] focus is[latex]\,\left(1,\frac{1}{3}\right).[/latex]

For the following exercises, determine the equation for the parabola from its graph.

[latex]{x}^{2}=y[/latex]

[latex]{\left(y-2\right)}^{2}=\frac{1}{4}\left(x+2\right)[/latex]

[latex]{\left(y-\sqrt{3}\right)}^{2}=4\sqrt{5}\left(x+\sqrt{2}\right)[/latex]

For the following exercises, the vertex and endpoints of the latus rectum of a parabola are given. Find the equation.

[latex]V\left(0,0\right),\text{Endpoints }\left(2,1\right),\left(-2,1\right)[/latex]

[latex]V\left(0,0\right),\text{Endpoints }\left(-2,4\right),\left(-2,-4\right)[/latex]

[latex]{y}^{2}=-8x[/latex]

[latex]V\left(1,2\right),\text{Endpoints }\left(-5,5\right),\left(7,5\right)[/latex]

[latex]V\left(-3,-1\right),\text{Endpoints }\left(0,5\right),\left(0,-7\right)[/latex]

[latex]{\left(y+1\right)}^{2}=12\left(x+3\right)[/latex]

[latex]V\left(4,-3\right),\text{Endpoints }\left(5,-\frac{7}{2}\right),\left(3,-\frac{7}{2}\right)[/latex]

Real-World Applications

The mirror in an automobile headlight has a parabolic cross-section with the light bulb at the focus. On a schematic, the equation of the parabola is given as[latex]\,{x}^{2}=4y.\,[/latex]At what coordinates should you place the light bulb?

[latex]\left(0,1\right)[/latex]

If we want to construct the mirror from the previous exercise such that the focus is located at[latex]\,\left(0,0.25\right),[/latex] what should the equation of the parabola be?

A satellite dish is shaped like a paraboloid of revolution. This means that it can be formed by rotating a parabola around its axis of symmetry. The receiver is to be located at the focus. If the dish is 12 feet across at its opening and 4 feet deep at its center, where should the receiver be placed?

At the point 2.25 feet above the vertex.

Consider the satellite dish from the previous exercise. If the dish is 8 feet across at the opening and 2 feet deep, where should we place the receiver?

A searchlight is shaped like a paraboloid of revolution. A light source is located 1 foot from the base along the axis of symmetry. If the opening of the searchlight is 3 feet across, find the depth.

0.5625 feet

If the searchlight from the previous exercise has the light source located 6 inches from the base along the axis of symmetry and the opening is 4 feet, find the depth.

An arch is in the shape of a parabola. It has a span of 100 feet and a maximum height of 20 feet. Find the equation of the parabola, and determine the height of the arch 40 feet from the center.

If the arch from the previous exercise has a span of 160 feet and a maximum height of 40 feet, find the equation of the parabola, and determine the distance from the center at which the height is 20 feet.

An object is projected so as to follow a parabolic path given by[latex]\,y=-{x}^{2}+96x,[/latex]where[latex]\,x\,[/latex]is the horizontal distance traveled in feet and[latex]\,y\,[/latex]is the height. Determine the maximum height the object reaches.

For the object from the previous exercise, assume the path followed is given by[latex]\,y=-0.5{x}^{2}+80x.\,[/latex] Determine how far along the horizontal the object traveled to reach maximum height.

• Algebra and Trigonometry. Authored by : Jay Abramson, et. al. Provided by : OpenStax CNX. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]

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Question 4 - Case Based Questions (MCQ) - Chapter 2 Class 10 Polynomials

Last updated at April 16, 2024 by Teachoo

Applications of Parabolas: Highway Overpasses/  Underpasses  A highway underpass is parabolic in shape.

Parabola a parabola is the graph that results from p(x) = ax 2 + bx + c. parabolas are symmetric about a vertical line known as the axis of symmetry. the axis of symmetry runs through the maximum or minimum point of the parabola which is called the vertex..

If the highway overpass is represented by x 2 – 2x – 8.  Then its zeroes are

(a) (2, – 4)  , (b) (4, – 2), (c) (– 2, – 2)  , (d) (– 4, – 4).

The highway overpass is represented graphically.  Zeroes of a polynomial can be expressed  graphically. Number of zeroes of polynomial  is equal to number of points where the graph of  polynomial:

(a) intersects x-axis, (b) intersects y-axis, (c) intersects y-axis or x-axis, (d) none of the above.

Graph of a quadratic polynomial is a:

(a) straight line  , (c) parabola  , (d) ellipse.

The representation of Highway Underpass whose one zero is 6 and sum of the zeroes is 0, is:

(a) x 2 – 6x + 2  , (b) x 2 – 36, (c) x 2 – 6 , (d) x 2 – 3.

The number of zeroes that polynomial f(x) = (x – 2) 2   + 4 can have is: 

(a) 1  , (c) 0  .

Question Applications of Parabolas: Highway Overpasses/ Underpasses A highway underpass is parabolic in shape. Shape of Cross Slope: Parabola A parabola is the graph that results from p(x) = ax2 + bx + c. Parabolas are symmetric about a vertical line known as the Axis of Symmetry. The Axis of Symmetry runs through the maximum or minimum point of the parabola which is called the vertex. Question 1 If the highway overpass is represented by x2 – 2x – 8. Then its zeroes are (a) (2, – 4) (b) (4, – 2) (c) (– 2, – 2) (d) (– 4, – 4) Finding Zeroes of polynomial x2 – 2x – 8 = 0 Using splitting the middle term x2 – 4x + 2x – 8 = 0 x(x – 4) + 2(x – 4) = 0 (x – 4) (x + 2) = 0 ∴ x = 4, x = –2 So, the correct answer is (B) Question 2 The highway overpass is represented graphically. Zeroes of a polynomial can be expressed graphically. Number of zeroes of polynomial is equal to number of points where the graph of polynomial: (a) Intersects X-axis (b) Intersects Y-axis (c) Intersects Y-axis or X-axis (d) None of the above We know that Number of zeroes is equal to number of times parabola intersects the x-axis So, the correct answer is (A) Question 3 Graph of a quadratic polynomial is a: (a) straight line (b) circle (c) parabola (d) ellipse Graph of a quadratic polynomial is a parabola. So, the correct answer is (C) Question 4 The representation of Highway Underpass whose one zero is 6 and sum of the zeroes is 0, is: (a) x2 – 6x + 2 (b) x2 – 36 (c) x2 – 6 (d) x2 – 3 Given a polynomial where One Zero = 𝛼 = 6 And, Sum of zeroes = 0 𝛼 + 𝛽 = 0 6 + 𝛽 = 0 𝛽 = −6 Since zeroes of polynomial are 6 and −6 Our quadratic polynomial is (x − 6) (x − (−6)) (x − 6) (x + 6) x2 − 62 x2 − 36 So, the correct answer is (B) Question 5 The number of zeroes that polynomial f(x) = (x – 2)2 + 4 can have is: (a) 1 (b) 2 (c) 0 (d) 3 Given f(x) = (x – 2)2 + 4 = x2 + 4 – 4x + 4 = x2 – 4x + 8 Finding Zeros of f(x) x2 – 4x + 8 Using quadratic formula a = 1, b = −4, c = 8 Finding D D = b2 −4ac = (−4)2 − 4 × 1 × 8 = 16 − 32 = −16 Since D < 0 It has no real roots Thus, f(x) does not have any zeroes ∴ Number of zeroes of f(x) = 0 So, the correct answer is (C)

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CASE STUDY QUESTION 13-Class X-Maths

• August 30, 2021
• Class 10 , Mathematics

Applications of Parabolas-Highway Overpasses/Underpasses. A highway underpass is parabolic in shape.

Parabola A parabola is the graph that results from p(x)=ax 2 + bx + c Parabolas are symmetric about a vertical line known as the Axis of Symmetry. The Axis of Symmetry runs through the maximum or minimum point of the parabola which is called the vertex.

(b) The highway overpass is represented graphically. Zeroes of a polynomial can be expressed graphically. Number of zeroes of polynomial is equal to number of points where the graph of polynomial . (i) Intersects x-axis (ii) Intersects y-axis (iii) Intersects y-axis or x-axis (iv)None of the above

Sol. (i) Intersects x-axis

(c) Graph of a quadratic polynomial is a (i) straight line (ii) circle (iii)parabola (iv)ellipse

Sol. (iii) parabola

(d) The representation of Highway Underpass whose one zero is 6 and sum of the zeroes is 0, is (i) x 2 – 6x + 2 (ii) x 2 – 36 (iii) x 2 – 6 (iv) x 2 – 3

(e) The number of zeroes that polynomial f(x) = (x – 2) 2 + 4 can have is: (i)1 (ii) 2 (iii) 0 (iv) 3

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10th Class Mathematics Polynomials Question Bank

Done case based (mcqs) - polynomials total questions - 40.

A) all are real numbers done clear

B) all are rational numbers done clear

C) 'a' is a non-zero real number and b and c are any real numbers done clear

D) all are integers done clear

question_answer 2) If the roots of the quadratic polynomial are equal, and the discriminant \[D={{b}^{2}}-4ac,\] then:

A) \[D>0\] done clear

B) \[D<0\]      done clear

C) \[D\ge 0\] done clear

D) \[D=0\] done clear

question_answer 3) If \[\alpha \] and \[\frac{1}{\alpha }\]are the zeroes of the quadratic polynomial \[2{{x}^{2}}-x+8k,\] then k is:

A) \[4\] done clear

B) \[\frac{1}{4}\] done clear

C) \[\frac{-1}{4}\] done clear

D) \[2\] done clear

question_answer 4) The graph of \[{{\text{x}}^{\text{2}}}+\text{4}=0:\]

A) intersects X-axis at two distinct points done clear

B) touches X-axis at a point done clear

C) neither touches nor intersects X-axis done clear

D) either touches or intersects X-axis. done clear

question_answer 5) If the sum of the roots is \[-p\] and product of the roots \[-\frac{1}{p}\] is, then the quadratic polynomial is:

A) \[k\left( -p{{x}^{2}}+\frac{x}{p}+1 \right)\] done clear

B) \[k\left( p{{x}^{2}}-\frac{x}{p}-1 \right)\] done clear

C) \[k\left( {{x}^{2}}+px-\frac{1}{p} \right)\] done clear

D) \[k\left( {{x}^{2}}-px+\frac{1}{p} \right)\] done clear

A) spiral done clear

B) ellipse done clear

C) linear done clear

D) parabola done clear

question_answer 7) The graph of parabola opens downwards, if..........

A) \[a\ge 0\] done clear

B) \[a=0\] done clear

C) \[a<0\] done clear

D) \[a>0\] done clear

A) 0 done clear

B) 1 done clear

C) 2 done clear

D) 3 done clear

question_answer 9) The quadratic polynomial of the two zeroes in the above shown graph are:

A) \[k({{x}^{2}}-2x-8)\] done clear

B) \[k({{x}^{2}}+2x-8)\] done clear

C) \[k({{x}^{2}}+2x+8)\] done clear

D) \[k({{x}^{2}}-2x+8)\] done clear

question_answer 10) The   zeroes   of   the   quadratic   polynomial \[4\sqrt{3}\,{{x}^{2}}+5x-2\sqrt{3}\] are:

A) \[\frac{2}{\sqrt{3}}.\frac{\sqrt{3}}{4}\] done clear

B) \[-\frac{2}{\sqrt{3}}.\frac{\sqrt{3}}{4}\] done clear

C) \[\frac{2}{\sqrt{3}}.-\frac{\sqrt{3}}{4}\] done clear

D) \[-\frac{2}{\sqrt{3}}.-\frac{\sqrt{3}}{4}\] done clear

A) \[(2,-4)\] done clear

B) \[(4,-2)\] done clear

C) \[(-2,-2)\] done clear

D) \[(-4,-4)\] done clear

A) Intersects X-axis done clear

B) Intersects V-axis done clear

C) Intersects /-axis or X-axis done clear

D) None of these done clear

question_answer 13) Graph of a quadratic polynomial is a:

A) straight line done clear

B) circle done clear

C) parabola done clear

D) ellipse done clear

question_answer 14) The representation of Highway Underpass whose one zero is 6 and sum of the zeroes is 0, is:

A) \[{{x}^{2}}-6x+2\] done clear

B) \[{{x}^{2}}-36\] done clear

C) \[{{x}^{2}}-6\] done clear

D) \[{{x}^{2}}-3\] done clear

question_answer 15) The number of zeroes that polynomial \[f(x)={{(x-2)}^{2}}+4\] can have is:

A) 1 done clear

B) 2 done clear

C) 0 done clear

A) Parabola done clear

B) Hyperbola done clear

C) Ellipse done clear

question_answer 17) The zeroes of given quadratic polynomial are:

A) \[2,-4\] done clear

B) \[-2,4\] done clear

C) \[3,-4\] done clear

D) \[-3,4\] done clear

question_answer 18) Read from the graph the value of y corresponding to \[x=-1\] is:

A) \[-8\] done clear

B) \[-6\] done clear

C) \[-5\] done clear

D) \[-2\] done clear

question_answer 19) The graph of the given quadratic polynomial cut at which points on the X-axis?

A) \[(-2,0),\,\,(4,0)\] done clear

B) \[(0,-2),\,\,(0,4)\] done clear

C) \[(0,-2),\,\,(0,-8)\] done clear

D) \[None\,\, of\,\, the\,\,above\] done clear

question_answer 20) The graph of the given quadratic polynomial cut at which point on X-axis?

A) \[(-8,0)\] done clear

B) \[(0,-8)\] done clear

C) \[(-10,0)\] done clear

D) \[(-10,0)\] done clear

C) 3 done clear

D) 4 done clear

question_answer 22) Which type of polynomial is represented by Jay's graph?

A) Linear done clear

B) Parabola done clear

C) Zig-zag done clear

question_answer 23) How many zeroes are there for the Richa's graph?

question_answer 24) If \[p(x)=a{{x}^{2}}+bx+c\] and \[a+b+c=0,\] then one zero is:

A) \[\frac{-b}{a}\]  done clear

B) \[\frac{c}{a}\] done clear

C) \[\frac{b}{c}\] done clear

D) \[-\frac{c}{a}\] done clear

question_answer 25) If \[p(x)=a{{x}^{2}}+bx+c\] and \[a+c=b,\] then one of the zeroes is:

A) \[\frac{b}{a}\] done clear

C) \[\frac{-c}{a}\] done clear

D) \[\frac{-b}{a}\] done clear

A) Yes done clear

B) No done clear

C) Can't say done clear

question_answer 27) If yes, then the correct quadratic polynomial is:

A) \[4\sqrt{3}{{x}^{2}}-5x+2\sqrt{3}\] done clear

B)        \[4\sqrt{3}{{x}^{2}}+5x-2\sqrt{3}\] done clear

C) \[4\sqrt{3}{{x}^{2}}+5x+2\sqrt{3}\] done clear

D) \[4\sqrt{3}{{x}^{2}}-5x-2\sqrt{3}\] done clear

question_answer 28) The value of \[{{\alpha }^{2}}+{{\beta }^{2}}\] is:

A) \[\frac{53}{48}\] done clear

B) \[\frac{59}{48}\] done clear

C) \[\frac{73}{48}\] done clear

D) \[\frac{71}{48}\] done clear

question_answer 29) What is the value of the correct polynomial if \[x=-1\]?

A) \[-5+2\sqrt{3}\] done clear

B) \[5-2\sqrt{3}\] done clear

C) \[5-6\sqrt{3}\] done clear

D) \[-5+6\sqrt{3}\] done clear

question_answer 30) The value of \[{{\alpha }^{3}}-{{\beta }^{3}}\]is:

A) \[\frac{-539}{192\sqrt{3}}\] done clear

B) \[\frac{539}{192\sqrt{3}}\] done clear

C) \[\frac{539\sqrt{3}}{192}\] done clear

D) \[\frac{-539\sqrt{3}}{192}\] done clear

A) \[\frac{3}{5}\] done clear

B) \[\frac{3}{4}\] done clear

C) \[\frac{2}{5}\] done clear

D) \[\frac{1}{5}\] done clear

question_answer 32) The zeroes of the quadratic polynomial \[{{x}^{2}}+20x+96\]are:

A) both positive done clear

B) both negative done clear

C) one positive and other negative done clear

D) both equal done clear

question_answer 33) The quadratic polynomial whose zeroes are 5 and \[-12\] is given by:

A) \[{{x}^{2}}+7x-60\] done clear

B) \[15{{x}^{2}}-x-6\] done clear

C) \[{{x}^{2}}-7x+60\] done clear

D) \[15{{x}^{2}}+x+6\] done clear

question_answer 34) If one zero of the polynomial \[f(x)=5{{x}^{2}}+13x+m\] is reciprocal of the other, then the value of m is:

A) 6 done clear

B) 0 done clear

C) 5 done clear

D) 1 done clear

question_answer 35) Which of the following cannot be the graph, of a quadratic polynomial?

question_answer 37) If the product of the zeroes of the quadratic polynomial \[f(x)=a{{x}^{2}}-6x-6\] is 4, then the value of a is:

A) \[\frac{-3}{2}\] done clear

B) \[\frac{3}{2}\] done clear

C) \[\frac{2}{3}\] done clear

D) \[\frac{-2}{3}\] done clear

question_answer 38) The flow of the water in the pool is represented by \[{{x}^{2}}-2x-8,\] then its zeroes are:

B) \[4,-2\] done clear

C) \[2,-2\] done clear

D) \[-4,-4\] done clear

question_answer 39) If a and p be the zeroes of the polynomial \[{{x}^{2}}-1,\]then the value of \[\frac{1}{\alpha }+\frac{1}{\beta }\]is:

A) \[0\] done clear

B) \[\frac{1}{2}\] done clear

C) \[1\] done clear

D) \[-1\] done clear

question_answer 40) A quadratic polynomial whose one zero is \[-3\] and product of zeroes is 0, is:

A) \[3{{x}^{2}}+3\] done clear

B) \[{{x}^{2}}-3x\] done clear

C) \[{{x}^{2}}+3x\] done clear

D) \[3{{x}^{2}}-3\] done clear

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Case Based (MCQs) - Polynomials

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Conics in Real Life

Table of Contents

27 October 2020                

Read time: 8 minutes

Introduction

Conics or conic sections were studied by Greek mathematicians, with Apollonius of Pergo’s work on their properties around 200 B.C.

Conics sections are planes, cut at varied angles from a cone. The shapes vary according to the angle at which it is cut from the cone.

As they are cut from cones, they are called Conies. Further, they have some common properties as they all belong to cones. These curved sections are related to.

Is Geometry hard?

Conics in Real life

Here is a PDF that tells us more about conics in real life. They can think of these.  Click on the download button to explore them.

What is Conic Section?

Conic section is a curve obtained by the intersection of the surface of a cone with a plane.

In Analytical Geometry, a conic is defined as a plane algebraic curve of degree 2. That is, it consists of a set of points which satisfy a quadratic equation in two variables. This quadratic equation may be written in matrix form. By this, some geometric properties can be studied as algebraic conditions.

Thus, by cutting and taking different slices(planes) at different angles to the edge of a cone, we can create a circle, an ellipse, a parabola, or a hyperbola, as given below

The circle is a type of ellipse, the other sections are non-circular. So, the circle is of fourth type.

Focus, Directrix and Eccentricity

The curve is also defined by using a point(focus) and a straight line (Directrix).

If we measure and let

a – the perpendicular distance from the focus to a point P on the curve,

and b – the distance from the directrix to the point P,

then a: b will always be constant.

\(a: b < 1\) for ellipse

\(a: b= 1\) for parabola as \(a= b\)

and \(a: b> 1\) for hyperbola.

Eccentricity: The above ratio a: b is the eccentricity.

Thus, any conic section has all the points on it such that the distance between the points to the focus is equal to the eccentricity times that of the directrix.

Thus, if eccentricity \(<1\) , it is an ellipse.

if eccentricity \(=1\) , it is a parabola.

and if eccentricity \(=1\) , it is a hyperbola.

For a circle, eccentricity is zero. With higher eccentricity, the conic is less curved.

Latus Rectum

The line parallel to the directrix and passing through the focus is Latus Rectum.

Length of Latus Rectum = 4 times the focal length

Length \(=\frac{2b^2}{a}\) where \(a =\frac{1}{2}\) the major diameter

and \(b =\frac{1}{2}\) the minor diameter.

= the diameter of the circle.

Ellipse has a focus and directrix on each side i.e., a pair of them.

General equation for all conics is with cartesian coordinates x and y and has \(x^2\)  and \(y^2\)  as

the section is curved. Further, x, y, x y and factors for these and a constant is involved.

Thus, the general equation for a conic is

\[Ax^2 + B x y + C y^2+ D x + E y + F = 0\]

Using this equation, following equations are obtained:

For ellipse, \(x^2a^2+y^2b^2=1\)

For hyperbola, \(x^2a^2-y^2b^2=1\)

For circle, \(x^2a^2+y^2a^2=1\) (as radius is a)

Parabola in Real Life

Parabola is obtained by slicing a cone parallel to the edge of the cone. It is of U – shape as a stretched geometric plane. This formula is \(y =x^2\) on the x – y axis.

Mathematician Menaechmus derived this formula.

Parabola is found in nature and in works of man.

Water from a fountain takes a path of parabola to fall on the earth.

A ball thrown high, follows a parabolic path.

A roller coaster takes the path of rise and fall of a parabolic track of the sea.

An architectural structure built and named The Parabola in London in 1962 has a copper roof with parabolic and hyperbolic linings.

The Golden Gate Bridge in San Francisco in California is famous with parabolic spans on both sides.

In light houses, parabolic bulbs are provided to have a good focus of beam to be seen from distance by mariners.

Automobile headlights are also with parabola type.

The stretched arc of a rocket launch is parabolic.

The satellite dish is a parabolic structure facilitating focus and reflection of radio waves.

Ellipses in Real Life

Electrons in the atom move around the nucleus in an elliptical path of orbit.

Property of Ellipse to reflect sound and light is used in pulverizing kidney stones. The patient is laid in an elliptical tank of water. Kidney stones being at the other focus are concentrated and pulverized.

Paul’s Cathedral is an elliptical shaped structure to facilitate talking at one end is heard at the other end using the property of ellipse.

There is an ellipse shaped park in front of White House in Washington.

When a tumbler of water is tilted, an elliptical surface of water is seen.

Food items carrot, cucumber cut at an angle to its main axis results in elliptical shape and elegant look.

Whispering galleries at US Statutory capital and St. Paul’s Cathedral, London demonstrates the property of the ellipse that one’s whisper from one focus can be heard at the other focus by only a person to whom it is sent  

Elliptical training machines enable running or walking without straining the heart.

Hyperbolas in Real Life

A guitar is an example of hyperbola as its sides form hyperbola.

Dulles Airport has a design of hyperbolic parabolic. It has one cross-section of a hyperbola and the other a parabola.

Gear Transmission having pair of hyperbolic gears. It is with skewed axles and hourglass shape giving hyperbola shape. The hyperbolic gears transmit motion to the skewed axle.

The Kobe Port Tower has hourglass shape, that means it has two hyperbolas. Things seen from a point on one side will be the same when seen from the same point on the other side.

Satellite systems, Radio systems use hyperbolic functions.

Inverse relationship is related to hyperbola. Pressure and Volume of gas are in inverse relationships. This can be described by a hyperbola.

Lens, monitors, and optical glasses are of hyperbola shape.

Conic or conical shapes are planes cut through a cone. Based on the angle of intersection, different conics are obtained. Parabola, Ellipse, and Hyperbola are conics. Circle is a special conic. Conical shapes are two dimensional, shown on the x, y axis. Conic shapes are widely seen in nature and in man-made works and structures. They are beneficially used in electronics, architecture, food and bakery and automobile and medical fields.

Frequently Asked Questions (FAQs)

What are the 4 types of conic sections.

According to the angle of intersection between a plane and a cone, four different conic sections are obtained. They are Parabola, Ellipse, Hyperbola, and Circle. They are two dimensional on the x-y axis.

How do we obtain Conic Sections?

Conic section involves a cutting plane, surface of a double cone in hourglass form and the intersection of the cone by the plane. According to the angle of cutting, that is, light angle, parallel to the edge and deep angle, ellipse, parabola and hyperbola respectively are obtained. Circle is also conic, and it is cut parallel to the circular bottom face of the cone.

How do you find the area of an ellipse?

Area of an ellipse is \((a \times b \times π)\) sq. units.

where a = length of major axis of ellipse

b = length of minor axis of ellipse.

What are some real-life applications of conics?

Planets travel around the Sun in elliptical routes at one focus.

Mirrors used to direct light beams at the focus of the parabola are parabolic.

Parabolic mirrors in solar ovens focus light beams for heating.

Sound waves are focused by parabolic microphones.

Car headlights and spotlights are designed based on parabola’s principles.

The path travelled by objects thrown into air is parabolic.

Hyperbolas are used in long range navigation systems called LORAN.

Telescopes use parabolic mirrors.

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Case Study Based- 3 Applications of Parabolas-Highway Overpasses/Underpasses A highway underpass is parabolic in shape. Parabola A parabola is the graph that results from p ( x ) = a x 2 + b x + c Parabolas are symmetric about a vertical line known as the Axis of Symmetry. The Axis of Symmetry runs through the maximum or minimum point of the parabola which is called the Vertex. The number of zeroes that polynomial f ( x ) = ( x − 2 ) 2 + 4 can have is:

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Step by step video & image solution for Case Study Based- 3 Applications of Parabolas-Highway Overpasses/Underpasses A highway underpass is parabolic in shape. Parabola A parabola is the graph that results from p(x) = ax^(2)+bx+c Parabolas are symmetric about a vertical line known as the Axis of Symmetry. The Axis of Symmetry runs through the maximum or minimum point of the parabola which is called the Vertex. The number of zeroes that polynomial f(x) = (x - 2)^(2) + 4 can have is: by Maths experts to help you in doubts & scoring excellent marks in Class 10 exams.

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The locus of the mid-point of the chords of the parabola y 2 = 4 a x which subtend a right angle at the vertex of the parabola, is.

Write the axis of symmetry of the parabola y 2 = x .

The point on the axis of symmetry that bisects the parabola is called the "vertex”, and it is the point where the curvature is greatest.

Is the parabola y 2 = 4 x symmetrical about x-axis ?

Parabolas having their vertices at the origin and foci on the x-axis.

The line that splits the parabola tiirough tiie'middle is called the "axis of symmetry".

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Case Study Based- 4 100m RACE A stopwatch was used to find the time...

Case Study Based- 4 100m RACE A stopwatch was used to find the tim...

Case Study Based- 3 Applications of Parabolas-Highway Overpasses/Underpasses A highway underpass is parabolic in shape. Parabola A parabola is the graph that results from p(x)=ax 2 +bx+c Parabolas are symmetric about a vertical line known as the Axis of Symmetry. The Axis of Symmetry runs through the maximum or minimum point of the parabola which is called the Vertex (a) If the highway overpass is represented by x 2 – 2x – 8. Then its zeroes are (i) (2,-4) (ii) (4,-2) (iii) (-2,-2) (iv) (-4,-4) (b) The highway overpass is represented graphically. Zeroes of a polynomial can be expressed graphically. Number of zeroes of polynomial is equal to number of points where the graph of polynomial (i) Intersects x-axis (ii) Intersects y-axis (iii) Intersects y-axis or x-axis (iv)None of the above (c) Graph of a quadratic polynomial is a (i) straight line (ii) circle (iii)parabola (iv)ellipse (d) The representation of Highway Underpass whose one zero is 6 and sum of the zeroes is 0, is (i) x 2 – 6x + 2 (ii) x 2 – 36 (iii)x 2 – 6 (iv)x 2 – 3 (e) The number of real zeroes that polynomial f(x) = (x – 2) 2 + 4 can have is: (i)1 (ii) 2 (iii) 0 (iv) 3

(a) - (ii) (4,-2)

(b) - (i) Intersects x-axis

(c) - (iii) parabola

(d) - (ii) x 2 – 36

(e) - (iii) 0