# 7.4: The Parabola - Mathematics

Learning Objectives

• 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.

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 (Figure (PageIndex{1})), 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

Previously, 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 (Figure (PageIndex{2})).

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 ((x,y)) 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.

Previously, we learned about a parabola’s vertex and axis of symmetry. Now we extend the discussion to include other key features of the parabola (Figure (PageIndex{3})). 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 distanced d from the focus to any point (P) on the parabola is equal to the distance from (P) to the directrix.

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 ((x,y)) be a point on the parabola with vertex ((0,0)), focus ((0,p)),and directrix (y=−p) as shown in Figure (PageIndex{4}). The distanced d from point ((x,y)) to point ((x,−p)) on the directrix is the difference of the y-values: (d=y+p). The distance from the focus ((0,p)) to the point ((x,y)) is also equal to (d) and can be expressed using the distance formula.

[ egin{align*} d &=sqrt{{(x−0)}^2+{(y−p)}^2} [4pt] &=sqrt{x^2+{(y−p)}^2} end{align*} ]

Set the two expressions for (d) equal to each other and solve for (y) to derive the equation of the parabola. We do this because the distance from ((x,y)) to ((0,p)) equals the distance from ((x,y)) to ((x,−p)).

[sqrt{x^2+{(y−p)}^2}=y+p ]

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

[ egin{align*} x^2+{(y−p)}^2 &={(y+p)}^2 [4pt] x^2+y^2−2py+p^2 &=y^2+2py+p^2 [4pt] x^2−2py &=2py [4pt] x^2 &=4py end{align*} ]

The equations of parabolas with vertex ((0,0)) are (y^2=4px) when the x-axis is the axis of symmetry and (x^2=4py) 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))

Table (PageIndex{1}) and Figure (PageIndex{5}) summarize the standard features of parabolas with a vertex at the origin.

Table (PageIndex{1})
Axis of SymmetryEquationFocusDirectrixEndpoints of Latus Rectum
x-axis(y^2=4px)((p, 0))(x=−p)((p, pm 2p))
y-axis(x^2=4py)((0, p))(y=−p)((pm 2p, p))

The key features of a parabola are its vertex, axis of symmetry, focus, directrix, and latus rectum (Figure (PageIndex{5})). 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 (PageIndex{6}).

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

1. Determine which of the standard forms applies to the given equation: (y^2=4px) or (x^2=4py).
2. Use the standard form identified in Step 1 to determine the axis of symmetry, focus, equation of the directrix, and endpoints of the latus rectum.
• If the equation is in the form (y^2=4px), then
• the axis of symmetry is the (x)-axis, (y=0)
• set (4p) equal to the coefficient of (x) in the given equation to solve for (p). If (p>0), the parabola opens right. If (p<0), the parabola opens left.
• use (p) to find the coordinates of the focus, ((p,0))
• use (p) to find the equation of the directrix, (x=−p)
• use (p) to find the endpoints of the latus rectum, ((p,pm 2p)). Alternately, substitute (x=p) into the original equation.
• If the equation is in the form (x^2=4py),then
• the axis of symmetry is the (y)-axis, (x=0)
• set (4p) equal to the coefficient of (y) in the given equation to solve for (p). If (p>0), the parabola opens up. If (p<0), the parabola opens down.
• use (p) to find the coordinates of the focus, ((0,p))
• use (p) to find equation of the directrix, (y=−p)
• use (p) to find the endpoints of the latus rectum, ((pm 2p,p))
3. Plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.

-axis as the Axis of Symmetry

Graph (y^2=24x). Identify and label the focus, directrix, and endpoints of the latus rectum.

Solution

The standard form that applies to the given equation is (y^2=4px). Thus, the axis of symmetry is the x-axis. It follows that:

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

Next we plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola (Figure (PageIndex{7})).

Exercise (PageIndex{1})

Graph (y^2=−16x). Identify and label the focus, directrix, and endpoints of the latus rectum.

• Focus: ((−4,0))
• Directrix: (x=4)
• Endpoints of the latus rectum: ((−4,pm 8))

-axis as the Axis of Symmetry

Graph (x^2=−6y). Identify and label the focus, directrix, and endpoints of the latus rectum.

Solution

The standard form that applies to the given equation is (x^2=4py). Thus, the axis of symmetry is the (y)-axis. It follows that:

• (−6=4p),so (p=−dfrac{3}{2}). Since (p<0), the parabola opens down.
• the coordinates of the focus are ((0,p)=(0,−dfrac{3}{2}))
• the equation of the directrix is (y=−p=dfrac{3}{2})
• the endpoints of the latus rectum can be found by substituting (y=dfrac{3}{2}) into the original equation, ((pm 3,−dfrac{3}{2}))

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

Exercise (PageIndex{2})

Graph (x^2=8y). Identify and label the focus, directrix, and endpoints of the latus rectum.

• Focus: ((0,2))
• Directrix: (y=−2)
• Endpoints of the latus rectum: ((pm 4,2)).

## 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.

How to: Given its focus and directrix, write the equation for a parabola in standard form

1. Determine whether the axis of symmetry is the (x)- or (y)-axis.
1. If the given coordinates of the focus have the form ((p,0)), then the axis of symmetry is the (x)-axis. Use the standard form (y^2=4px).
2. If the given coordinates of the focus have the form ((0,p)), then the axis of symmetry is the (y)-axis. Use the standard form (x^2=4py).
2. Multiply (4p).
3. Substitute the value from Step 2 into the equation determined in Step 1.

Example (PageIndex{3}): Writing the Equation of a Parabola in Standard Form Given its Focus and Directrix

What is the equation for the parabola with focus ((−dfrac{1}{2},0)) and directrix (x=dfrac{1}{2})?

Solution

The focus has the form ((p,0)), so the equation will have the form (y^2=4px).

• Multiplying (4p), we have (4p=4(−dfrac{1}{2})=−2).
• Substituting for (4p), we have (y^2=4px=−2x).=

Therefore, the equation for the parabola is (y^2=−2x).

Exercise (PageIndex{3})

What is the equation for the parabola with focus (left(0,dfrac{7}{2} ight)) and directrix (y=−dfrac{7}{2})?

(x^2=14y).

## 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 (h) units horizontally and (k) units vertically, the vertex will be ((h,k)). This translation results in the standard form of the equation we saw previously with (x) replaced by ((x−h)) and (y) replaced by ((y−k)).

To graph parabolas with a vertex ((h,k)) other than the origin, we use the standard form ({(y−k)}^2=4p(x−h)) for parabolas that have an axis of symmetry parallel to the (x)-axis, and ({(x−h)}^2=4p(y−k)) 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))

Table (PageIndex{2}) and Figure (PageIndex{11}) summarize the standard features of parabolas with a vertex at a point ((h,k)).

Table (PageIndex{2})
Axis of SymmetryEquationFocusDirectrixEndpoints of Latus Rectum
(y=k)({(y−k)}^2=4p(x−h))((h+p, k))(x=h−p)((h+p, kpm 2p))
(x=h)({(x−h)}^2=4p(y−k))((h, k+p))(y=k−p)((hpm 2p, k+p))

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

1. Determine which of the standard forms applies to the given equation: ({(y−k)}^2=4p(x−h)) or ({(x−h)}^2=4p(y−k)).
2. Use the standard form identified in Step 1 to determine the vertex, axis of symmetry, focus, equation of the directrix, and endpoints of the latus rectum.
• If the equation is in the form ({(y−k)}^2=4p(x−h)),then:
• use the given equation to identify (h) and (k) for the vertex, ((h,k))
• use the value of (k) to determine the axis of symmetry, (y=k)
• set (4p) equal to the coefficient of ((x−h)) in the given equation to solve for (p). If (p>0),the parabola opens right. If (p<0), the parabola opens left.
• use (h), (k), and (p) to find the coordinates of the focus, ((h+p, k))
• use (h) andp p to find the equation of the directrix, (x=h−p)
• use (h), (k), and (p) to find the endpoints of the latus rectum, ((h+p,kpm 2p))
• If the equation is in the form ({(x−h)}^2=4p(y−k)),then:
• use the given equation to identify (h) and (k) for the vertex, ((h,k))
• use the value of (h) to determine the axis of symmetry, (x=h)
• set (4p) equal to the coefficient of ((y−k)) in the given equation to solve for (p). If (p>0), the parabola opens up. If (p<0), the parabola opens down.
• use (h), (k), and (p) to find the coordinates of the focus, ((h, k+p))
• use (k) and (p) to find the equation of the directrix, (y=k−p)
• use (h), (k), and (p) to find the endpoints of the latus rectum, ((hpm 2p, k+p))
3. Plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.

Example (PageIndex{4}): Graphing a Parabola with Vertex ((h, k)) and Axis of Symmetry Parallel to the (x)-axis

Graph ({(y−1)}^2=−16(x+3)). Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.

Solution

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

• the vertex is ((h,k)=(−3,1))
• the axis of symmetry is (y=k=1)
• (−16=4p),so (p=−4). Since (p<0), the parabola opens left.
• the coordinates of the focus are ((h+p,k)=(−3+(−4),1)=(−7,1))
• the equation of the directrix is (x=h−p=−3−(−4)=1)
• the endpoints of the latus rectum are ((h+p,kpm 2p)=(−3+(−4),1pm 2(−4))), or ((−7,−7)) and ((−7,9))

Next we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola (Figure (PageIndex{10})).

Exercise (PageIndex{4})

Graph ({(y+1)}^2=4(x−8)). Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.

• Vertex: ((8,−1))
• Axis of symmetry: (y=−1)
• Focus: ((9,−1))
• Directrix: (x=7)
• Endpoints of the latus rectum: ((9,−3)) and ((9,1)).

Example (PageIndex{5}): Graphing a Parabola from an Equation Given in General Form

Graph (x^2−8x−28y−208=0). Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.

Solution

Start by writing the equation of the parabola in standard form. The standard form that applies to the given equation is ({(x−h)}^2=4p(y−k)). 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 (x) in order to complete the square.

[ egin{align*} x^2−8x−28y−208&=0 [4pt] x^2−8x &=28y+208 [4pt] x^2−8x+16 &=28y+208+16 [4pt] (x−4)^2 &=28y+224 [4pt] (x−4)^2 &=28(y+8) [4pt] (x−4)^2&= 4⋅7⋅(y+8) end{align*}]

It follows that:

• the vertex is ((h,k)=(4,−8))
• the axis of symmetry is (x=h=4)
• since (p=7), (p>0) and so the parabola opens up
• the coordinates of the focus are ((h,k+p)=(4,−8+7)=(4,−1))
• the equation of the directrix is (y=k−p=−8−7=−15)
• the endpoints of the latus rectum are ((hpm 2p,k+p)=(4pm 2(7),−8+7)), or ((−10,−1)) and ((18,−1))

Next we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola (Figure (PageIndex{14})).

Exercise (PageIndex{5})

Graph ({(x+2)}^2=−20(y−3)). Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.

• Vertex: ((−2,3))
• Axis of symmetry: (x=−2)
• Focus: ((−2,−2))
• Directrix: (y=8)
• Endpoints of the latus rectum: ((−12,−2)) and ((8,−2)).

## 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 (Figure (PageIndex{16})). This is why the Olympic torch is ignited when it is held at the focus of the parabolic mirror.

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.

Example (PageIndex{6}): Solving Applied Problems Involving Parabolas

A cross-section of a design for a travel-sized solar fire starter is shown in Figure (PageIndex{17}). 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.

1. 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.
2. Use the equation found in part (a) to find the depth of the fire starter.

Solution

1. The vertex of the dish is the origin of the coordinate plane, so the parabola will take the standard form (x^2=4py),where (p>0). The igniter, which is the focus, is (1.7) inches above the vertex of the dish. Thus we have (p=1.7).

[egin{align*} x^2&=4pyqquad ext{Standard form of upward-facing parabola with vertex } (0,0) x^2&=4(1.7)yqquad ext{Substitute } 1.7 ext{ for } p x^2&=6.8yqquad ext{Multiply.} end{align*}]

1. The dish extends (dfrac{4.5}{2}=2.25) inches on either side of the origin. We can substitute (2.25) for (x) in the equation from part (a) to find the depth of the dish.

[egin{align*} x^2&=6.8yqquad ext{ Equation found in part } (a) {(2.25)}^2&=6.8yqquad ext{Substitute } 2.25 ext{ for } x y&approx 0.74qquad ext{Solve for } y end{align*}]

The dish is about (0.74) inches deep.

Exercise (PageIndex{6})

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.

1. 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).
2. Use the equation found in part (a) to find the depth of the cooker.

(y^2=1280x)

The depth of the cooker is (500) mm

## Key Equations

 Parabola, vertex at origin, axis of symmetry on x-axis (y^2=4px) Parabola, vertex at origin, axis of symmetry on y-axis (x^2=4py) Parabola, vertex at ((h,k)),axis of symmetry on x-axis ({(y−k)}^2=4p(x−h)) Parabola, vertex at ((h,k)),axis of symmetry on y-axis ({(x−h)}^2=4p(y−k))

## Key Concepts

• A parabola is the set of all points ((x,y)) 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 ((0,0)) and the x-axis as its axis of symmetry can be used to graph the parabola. If (p>0), the parabola opens right. If (p<0), the parabola opens left. See Example (PageIndex{1}).
• The standard form of a parabola with vertex ((0,0)) and the y-axis as its axis of symmetry can be used to graph the parabola. If (p>0), the parabola opens up. If (p<0), the parabola opens down. See Example (PageIndex{2}).
• When given the focus and directrix of a parabola, we can write its equation in standard form. See Example (PageIndex{3}).
• The standard form of a parabola with vertex ((h,k)) and axis of symmetry parallel to the (x)-axis can be used to graph the parabola. If (p<0), the parabola opens left. See Example (PageIndex{4}).
• The standard form of a parabola with vertex ((h,k)) and axis of symmetry parallel to the (y)-axis can be used to graph the parabola. See Example (PageIndex{5}).
• 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 Example (PageIndex{6}).

## Understanding Parabolas

Notice that the parabola a line of symmetry, meaning the two sides mirror each other.

There are two patterns for a parabola, as it can be either vertical (opens up or down) or horizontal (opens left or right.)

Patterns:

Let's look at a few key points about these patterns:

• If the x is squared, the parabola is vertical (opens up or down). If the y is squared, it is horizontal (opens left or right).
• If a is positive, the parabola opens up or to the right. If it is negative, it opens down or to the left.
• The vertex is at (h, k). You have to be very careful. Notice how the location of h and k switches based on if the parabola is vertical or horizontal. Also, the coordinate inside the parenthesis is negative, but the one outside is not.

Let's look at a couple parabolas and see what we can determine about them.

1.

First, we know that this parabola is vertical (opens either up or down) because the x is squared. We can determine it opens down because the a (-2) is negative.

Next we can find the vertex (h, k). For a vertical parabola, h is inside parenthesis, and since there is a negative in the pattern, we must take the opposite. So h = -3. k is outside, and the sign in the pattern is positive, so we will keep this number as is. k = 4. Thus, our vertex is (-3, 4).

Summary: This is a vertical parabola that opens down. Its vertex is (-3, 4).

2.

First, we know that this parabola is horizontal (opens either left or right) because the y is squared. We can determine it opens to the right because the a (1/2) is positive.

Next we can find the vertex (h, k). For a horizontal parabola, h is outside parenthesis, and since there is a positive in the pattern, we will leave it as is. So h = -1. k is inside, and the sign in the pattern is negative, so we will take the opposite. k = 4. Thus, our vertex is (-1, 4).

Summary: This is a horizontal parabola that opens to the right. Its vertex is (-1, 4).

Practice: Determine whether the parabola opens up, down, left, or right. Then find its vertex.

1.

2.

3.

4.

5.

Answers: 1) Opens up, vertex: (-5, -2) 2) Opens to the left, vertex: (-5, 4) 3) Opens to the right, vertex: (6, -3) 4) Opens up, vertex: (0, -3) 5) Opens down, vertex: (4, 1)

## Calculating the focus and directrix

Below is an example of how to calculate the focus and directrix that may provide a better understanding of the mathematical definition of a parabola provided above:

The focus is a point located on the same line as the axis of symmetry, while the directrix is a line perpendicular to the axis of symmetry. For parabolas, the focus is always on the inside of the parabola, and the directrix never touches the parabola. Because the vertex is the same distance from the focus and directrix, the directrix has a location directly opposite of the focus.

For a parabola in the vertex form y = a(x - h) 2 + k , the focus is located at (h, k + ) and the directrix is located at y = k - .

For horizontal parabolas, the vertex is x = a(y - k) 2 + h , where (h,k) is the vertex. The focus of parabolas in this form have a focus located at ( h + , k ) and a directrix at x = h - . The axis of symmetry is located at y = k .

## Talk:Parabola

Really nice job with the edits. Your overall presentation is exceptionally clear.

If it's possible to create a graph for the "Finding the Focus and Directrix" section, great, but if it doesn't work, no problem. One thing you could do to simplify this would be to graph a parabola, then add a dot and a dashed line using Microsoft word (Brendan could show you how to do this). Then, instead of actually putting mathematical expressions in the drawing itself, simply refer to the drawings in your text.

It would be really nice to add a section relating the material from the more mathematical section to the idea of a parabola as representing the trajectory of a thrown object, such as an interpretation of a, b, and c (standard form), and h and k (vertex form) for a thrown object (not so hard). This is totally optional, though.

The remaining changes I would say are necessary are all really small:

• At the beginning of the standard form section, maybe replace "The most commonly used. " with "Any vertically oriented parabola can be written using the equation [insert equation]. Also, the graph of any equation in this form will be a parabola." The problem with "most commonly used" is that it doesn't make it clear that a vertical parabola can always be described this way.
• To go from standard to vertex form, it requires that we factor the equation. In some cases, the equation as it is will be factorable. Factoring in general is only very indirectly helpful in writing a quadratic in standard form. For instance x^2 + 7x + 10 = (x+5)(x+2), but it takes a bit of doing to get from there to vertex form, so maybe delete these two sentences or replace them with something vague, like "sometimes there are short cuts, but in general.
• Two small issues with "Graphs of parabolas are often oriented vertically, so that the parabola opens upward or downward, though they can also open sideways. In this case, the equations for the parabola will be written oriented vertically to maintain consistency. To see horizontal parabolas, the x and y can be switched."
• Parabolas can actually also open diagonally. The equations just get much more unpleasant.
• The sentence "In this case. " is a really helpful sentence, the wording is just a bit confusing. By "In this case" do you mean "In the equations below" and when you say "the equations. will be written oriented vertically" I think you mean the equations represented vertically oriented parabolas, while the wording implies that the equations themselves are oriented vertically.

### Tanya 7/7

I believe I addressed most of the changes. I will try to make an image for the focus/directrix section, but I am not very confident in my drawing/graphing abilities.

### Abram 7/7

This page is really nice. I agree with Anna that it would help to hide each of the derivation steps.

The one other change that I think is critical (but it's much easier!) is to make all the y's in the more mathematical description lower case.

A couple of other small ideas, with most important changes at the beginning (but none of them are critical):

• At the beginning of your "Standard Form" section, qualify what types of parabolas this equation can apply to. Something like, "Graphs of parabolas are often oriented vertically, so that the parabola opens upward or downward. In this case, the equation of the parabola can be written as. "
• Similarly, in the derivation, explicitly note that this derivation is assuming a horizontal directrix, which is not necessary for parabolas in general.
• An image for the section "Finding the Focus and Directrix" would be extremely helpful, though I can imagine it might be a big pain to find or generate.
• In the Basic Description, eliminate the sentence that reads, "To the left we can see a plane slicing the lower cone so that it goes through the circular base." This sentence isn't really necessary, and eliminating it removes the need to talk about a finite cone two sentences after you specifically make the point that cones are infinite.

### Anna 7/6

I know it'd tedious to do, but I do think it would look better if you hid each of the derivation steps (I think one of Brendan's pages has a good example of this) of the derivation instead of the whole thing.

As someone clicking on it, my thoughts are "oh, let's see a derivation" *click* "GAHHH, that's so much math at once! How can I make it go away again?" Basically, having all of it appear at once is scary, and even my eyes glaze over it a bit. That's the only thing I'd change, though.

### Anna 7/4

I took out the word "general" in the mouse over for "base" to make that sentence more clear.

Can you hide the contents of the algebraic steps, while still keeping the summaries above? Like "now we square both sides" [show equation] ?

#### Gene 6/29

"but also in MANY OTHER FIELDS SUCH AS physics and engineering.'

### Anna 6/29

I wouldn't say that the parabola is "used in" those fields, I'd say it arises in them.

I'd force all of your equations to be the same size in your derivation. (you can do this by using left( and ight) for your parentheses for just one set of parentheses.)

Your mouse over on the word "vertex" should be reworded.

Change "like with standard form" to "as with standard form"

Are you going to finish up that last section here? Or do you think you might change it into an idea for the future?

Also, I bet that there are some really cool optics pictures and applets that illustrate the things you're talking about. You might want to google "parabolic mirror" or "parabolic lens" applets or images to get some more ideas and see new stuff. That would also bring you back to your initial idea that a parabola comes up in other fields.

## Questions

ROOTS
When you solve the quadratic equation y = x 2 &ndash 6x + 8 = 0 by factorization or by formula, x = 2 or 4. These are the 2 roots of the equation. What do you notice about these roots and the points at which the parabola crosses the x-axis?

TURNING POINT
The formula to find the x value of the turning point of the parabola is x = &ndashb/2a. Use this formula to find the x value where the graph turns. Substitute this x value into the equation y = x 2 &ndash 6x + 8 to find the y value of the turning point. What do you notice?

## Using the definition of a parabola to find the equation of a parabola.

Write an equation for a graph that is the set of all points in the plane that are at the same distance from the point (0,4) and the line y = -4

According to the definition, FG = GQ. Therefore, let us find all points P(x,y) such that FG and the distance from G to the line are equal.

## Real World Parabolas

The appearances of parabolic shapes in the physical world are very abundant. Here are a few examples with which you may be familiar. Below you will see five real-world structures, some of which are parabolas: a roller coaster, the reflector from a flashlight, the base of the Eiffel Tower, the McDonald's arches, and the flight trajectory of NASA's zero-G simulator, known as the "vomit comet".

Have you ever wondered why the best roller coasters are parabolic? When you're riding these coasters it feels like you're defeating the force of gravity, right? Exactly! When a coaster falls from the peak of the parabola, it is rejecting air resistance and all the bodies are falling at the same rate. The only force here is gravity.

G forces also play a role in your feeling of weightlessness and heaviness when falling. First, it is important to distinguish that gravity is not a G force. G force is measured in what you feel while sitting still in the earth's gravitational field. So, for example, when you are seated in a coaster, the seat is exerting the same amount of force as the earth, but it is directed oppositely, this is what keeps you sitting still. But, when you're falling you aren't experiencing any G's (zero G force) so actually the seat isn't supporting you at all! But once you are at the bottom of a drop, the G force returns to being greater than 1:

• G forces are greatest at the minimums of a parabola
• G forces are least at the maximum of a parabola

So, the shape of a coaster as well as the ascent and descent play a vital role in the rider's enjoyment. Parabolic-shaped coasters are enjoyed so much because of the intense pull of gravity and the nonexistent G force that occurs when falling.

## Equation of Circle and Parabolas

&emsp&emspA circle C in the XY plane, with center at the point (h, k) and radius r , is the set of all points at distance r from the point (h, k) . Let P : (x,y) be any point on C . Then by the distance formula from Section 7.1 we have

An equivalent equation is

Obtain an equation for the circle having its center at (-1,3/2) and a radius of 2 .
Substituting into (1) we obtain

This is a convenient form in which to leave the equation, since it readily yields the radius and the coordinates of the center.
&emsp&emspConsider the equation

We recall from Section 6.7 that by using the method of completing the square, this equation can be written in the form

We recognize from the form of (2) that its graph, and hence the graph of (1), is the circle with center (h, k) and radius r .

Completing the square we have

From this last equation we recognize that its graph is the circle with center (3/4,-1) and radius 7/4 .

Let's see how our graph generator solves this problem and similar problems and generate graphs. Click on "Solve Similar" button to see more examples.

7.6&emsp&emspSome Parabolas and Their Equations

&emsp&emspIn many situations a certain type of curve, called a parabola, appears. For example, the path traced out by a stone thrown into the air (not vertically) is a part of a parabola. The arc of water from a hose is a part of a parabola. The reflecting mirror in a car headlight is in the shape of a parabolic dish, and so are the mirrors in a good reﬂecting telescope.
In a coordinate plane, parabolas are graphs of certain types of second degree equations in the variables x and y . An example of such an equation is
(1)&emsp&emsp&emsp y=x^2
Plotting points and graphing we obtain the curve as shown in Figure 8. From the equation we can tell that as x increases positively or negatively y increases positively. The point at which the parabola turns most sharply is its vertex. The vertex of the parabola in Figure 8 is the point (0,0) . This parabola opens upward and the vertical line through its vertex divides it into mirror image halves. This line is called the axis of symmetry of the parabola.
The parabola y=x^2 is just one of the many parabolas with vertical axis and vertex at the origin. In fact, these parabolas are the graphs of equations of the form
(2)&emsp&emsp&emsp y=ax^2
where a is a nonzero real number. If a is positive, then the parabola opens upward. while if a is negative it opens down. To sketch the

 x y -3 9 -2 4 -1 1 0 0 1 1 2 4 3 9

graph of a parabola it is usually sufficient to plot the vertex and two or three points on each side of the axis of symmetry.

Example 1&emsp&emspGraph y=-(1/2)x^2 .

This is a parabola with vertex at the origin and the Y axis as its axis of symmetry. Since the coefficient -1/2 is negative, the parabola opens downward.

 x y -4 -8 -3 -9/2 -2 -2 -1 -1/2 0 0 1 -1/2 2 -2 3 -9/2 4 -8

also have graphs that are parabolas. In fact, they are parabolas with vertex at the point (h,k) and whose axis of symmetry is the vertical line through (h,k) . They open up or down depending on whether a is positive or negative.

Example 2.&emsp&emspGraph y-3=1/3(x+1)^2 .

This is a parabola with vertex at (-1,3) which opens up. Its axis of symmetry is the line x=-1 .

 x y -4 6 -3 13/3 -2 10/3 -1 3 0 10/3 1 13/3 2 6 3 25/3

Consider an equation of the form

By completing the square we can transform this equation to one of the form (3) above. Thus the graph of (4) is a parabola. From the resulting equation we can determine the vertex, axis of symmetry, and the direction in which the parabola opens.

Example 3.&emsp&emspGraph y=2x^2+3x-1 .

We complete the square on the right-hand side

The graph of the resulting equation is seen to be a parabola opening upward with vertex at (-3/4,-17/8) and axis of symmetry the line x=-3/4

 x y -2 1 -3 8 1 4 2 13

Similarly, equations of the form

have parabolas for graphs. These parabolas have horizontal axes of symmetry and open either to the left or to the right depending on whether a is negative or positive. Again we ﬁnd the vertex and axis of symmetry by completing the square.

The graph of this equation is a parabola with vertex at (1,2) , it opens
to the right, and its axis of symmetry is the line y=2 .

Let's see how our graph generator solves this problem and similar problems and generate graphs. Click on "Solve Similar" button to see more examples.

Find an equation of the parabola having vertex (3,1) , vertical axis of symmetry, and passing through the point P : (2,3) .

## Examples of Parabolas

Example 1: Find an equation of the parabola having its focus at $left( <0, – 3> ight)$ and as its directrix on the line $y = 3$.

Solution: Since the focus is on the Y-axis and is also below the directrix, the parabola opens downward, and $a = – 3$. Hence the equation of the parabola is $= – 12y$. The length of the latus rectum is $|4left( < – 3> ight)| = 12$.

Example 2: Given the parabola having the equation $= 7x$, find the coordinates of the focus, the equation of the directrix, and the length of the latus rectum.

Solution: Compared with the general equation, here we have $4a = 7 Rightarrow a = frac<7><4>$. Since $a > 0$, the parabola opens to the right. The focus is at the point $Fleft( <4>,0> ight)$.
The equation of the directrix is $x = – frac<7><4>$. The length of the latus rectum is $7$.

Example 3: Show that the ordinate at any point $P$ of the parabola is a mean proportional between the length of the latus rectum and the abscissa of $P$.

Solution: Let $Pleft( ight)$ be any point of the parabola
$= 4ax$

Then the length of the latus rectum is $l = 4a$, therefore from the above parabola equation:
[egin 4ax = Rightarrow left( ight)left( ight) = ight)^2> end ]

This shows that the ordinate at any point $P$ of the parabola is a mean proportional between the length of the latus rectum and the abscissa of $P$.

## Parabola : Standard equation of Parabola , Latus Rectum

DEFINITION : CONIC SECTION
A conic section or conic is the locus of a point, which moves in a plane so that its distance from a fixed point is in a constant ratio to its distance from a fixed straight line, not passing through the fixed point.

∎ The fixed point is called the focus.

∎ The fixed straight line is called the directrix.

∎ The constant ratio is called the eccentricity and is denoted by e.

∎ When the eccentricity is unity i.e., e = 1 , the conic is called a parabola

when e < 1, the conic is called an ellipse .

and when e > 1, the conic is called a hyperbola.

∎ The straight line passing through the focus and perpendicular to the directrix is called the axis of the parabola.

∎ A point of intersection of a conic with its axis is called vertex.

### Standard equation of a Parabola:

Let S be the focus, ZM the directrix and P the moving point. Draw SZ perpendicular from S on the directrix. Then SZ is the axis of the parabola. Now the middle point of SZ , say A , will lie on the locus of P ,

i.e., AS = AZ
Take A as the origin, the x-axis along AS, and the y-axis along the perpendicular to AS at A, as in the figure.

Let AS = a , so that ZA is also a .

Let (x, y) be the coordinates of the moving point P.

Then MP = ZN = ZA + AN = a + x .

So that , (a + x) 2 = (x – a) 2 + y 2

Hence , the equation of parabola is y 2 = 4ax

### Latus Rectum:

The chord of a parabola through the focus and perpendicular to the axis is called the latus rectum.

In the figure LSL’ is the latus rectum.

Also LSL’ = 4a = double ordinate through the focus S.

Note:
∎ Any chord of the parabola y 2 = 4ax perpendicular to the axis of the parabola is called double ordinate.
∎ Two parabolas are said to be equal when their latus recta are equal.

Four common forms of a Parabola

Four common forms of a Parabola:

∎ Equation of the Directrix: x = −a

∎ Equation of the Directrix : x = a

∎ Equation of the Directrix : y = -a

∎ Equation of the Directrix : y = a

Illustration : Find the vertex , axis , directrix, focus, latus rectum and the tangent at vertex for the parabola 9y 2 − 16x − 12y − 57 = 0

Solution: The given equation can be rewritten as

which is of the form Y 2 = 4AX.

Hence the vertex is (− 61/16 , 2/3)

The directrix is : X + A = 0

The focus is X = A and Y = 0

Length of the latus rectum = 4A = 16/9

The tangent at the vertex is X = 0

Illustration : The extreme points of the latus rectum of a parabola are (7, 5) and (7, 3). Find the equation of the parabola and the points where it meets the axes.

Solution: Focus of the parabola is the mid-point of the latus rectum.

Also axis of the parabola is perpendicular to the latus rectum and passes through the focus. Its equation is

Length of the latus rectum = (5 − 3) = 2

Hence the vertex of the parabola is at a distance 2/4 = 0.5 from the focus. We have two parabolas, one concave rightwards and the other concave leftwards.

The vertex of the first parabola is (6.5, 4)

and its equation is (y − 4) 2 = 2(x − 6.5)

and it meets the x-axis at (14.5 , 0).

The equation of the second parabola is

It meets the x-axis at (−0.5, 0) and the y-axis at (0, 4 ± √15).

Exercise :
(i) Find the equation of parabola whose focus is (1, −1) and whose vertex is (2, 1). Also find its axis and latus rectum.

(ii) Find vertex, focus, directix and latus rectum of the parabola y 2 + 4x + 4y − 3 = 0.

(iii) Find the equation of the parabola whose axis is parallel to the y – axis and which passes through the points (0, 4), (1, 9) and (−2, 6) and determine its latus rectum.