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The Parabola

By the end of this section, you will be able to:

  • 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

Katherine Johnson is the pioneering NASA mathematician who was integral to the successful and safe flight and return of many human missions as well as satellites. Prior to the work featured in the movie Hidden Figures, she had already made major contributions to the space program. She provided trajectory analysis for the Mercury mission, in which Alan Shepard became the first American to reach space, and she and engineer Ted Sopinski authored a monumental paper regarding placing an object in a precise orbital position and having it return safely to Earth. Many of the orbits she determined were made up of parabolas, and her ability to combine different types of math enabled an unprecedented level of precision. Johnson said, “You tell me when you want it and where you want it to land, and I’ll do it backwards and tell you when to take off.”

Johnson’s work on parabolic orbits and other complex mathematics resulted in successful orbits, Moon landings, and the development of the Space Shuttle program. Applications of parabolas are also critical to other areas of science. 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.

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)(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.

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. 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 dd from the focus to any point PP on the parabola is equal to the distance from PP 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)(x,y) be a point on the parabola with vertex (0,0)(0,0), focus (0,p)(0,p), and directrix y=py=-p. The distance dd from point (x,y)(x,y) to point (x,p)(x,-p) on the directrix is the difference of the yy-values: d=y+pd=y+p. The distance from the focus (0,p)(0,p) to the point (x,y)(x,y) is also equal to dd and can be expressed using the distance formula.

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

d=(x0)2+(yp)2=x2+(yp)2d=\sqrt{(x-0)^2+(y-p)^2}=\sqrt{x^2+(y-p)^2}x2+(yp)2=y+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.

x2+(yp)2=(y+p)2x2+y22py+p2=y2+2py+p2x22py=2pyx2=4py \begin{array}{lrcl} & 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{array}

The equations of parabolas with vertex (0,0)(0,0) are y2=4pxy^2=4px when the xx-axis is the axis of symmetry and x2=4pyx^2=4py when the yy-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). The table and figures below summarize the standard features of parabolas with a vertex at the origin.

Axis of SymmetryEquationFocusDirectrixEndpoints of Latus Rectum
xx-axisy2=4pxy^2=4px(p,0)(p,0)x=px=-p(p,±2p)(p,\pm2p)
yy-axisx2=4pyx^2=4py(0,p)(0,p)y=py=-p(±2p,p)(\pm2p,p)

(a) When p>0p>0 and the axis of symmetry is the xx-axis, the parabola opens right.

(b) When p<0p<0 and the axis of symmetry is the xx-axis, the parabola opens left.

(c) When p>0p>0 and the axis of symmetry is the yy-axis, the parabola opens up.

(d) When p<0p<0 and the axis of symmetry is the yy-axis, the parabola opens down.

The key features of a parabola are its vertex, axis of symmetry, focus, directrix, and latus rectum. 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 below.

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: y2=4pxy^2=4px or x2=4pyx^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 y2=4pxy^2=4px, then:
      • the axis of symmetry is the xx-axis, y=0y=0
      • set 4p4p equal to the coefficient of xx in the given equation to solve for pp. If p>0p>0, the parabola opens right. If p<0p<0, the parabola opens left.
      • use pp to find the coordinates of the focus, (p,0)(p,0)
      • use pp to find the equation of the directrix, x=px=-p
      • use pp to find the endpoints of the latus rectum, (p,±2p)(p,\pm2p). Alternately, substitute x=px=p into the original equation.
    • If the equation is in the form x2=4pyx^2=4py, then:
      • the axis of symmetry is the yy-axis, x=0x=0
      • set 4p4p equal to the coefficient of yy in the given equation to solve for pp. If p>0p>0, the parabola opens up. If p<0p<0, the parabola opens down.
      • use pp to find the coordinates of the focus, (0,p)(0,p)
      • use pp to find the equation of the directrix, y=py=-p
      • use pp to find the endpoints of the latus rectum, (±2p,p)(\pm2p,p)
  3. Plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.

Example. Graph y2=24xy^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 y2=4pxy^2=4px. Thus, the axis of symmetry is the xx-axis. It follows that:

  • 24=4p24=4p, so p=6p=6. Since p>0p>0, the parabola opens right.
  • the coordinates of the focus are (p,0)=(6,0)(p,0)=(6,0)
  • the equation of the directrix is x=p=6x=-p=-6
  • the endpoints of the latus rectum have the same xx-coordinate at the focus. To find the endpoints, substitute x=6x=6 into the original equation: (6,±12)(6,\pm12)

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

Graphy2=16xy^2=-16x. Give the coordinates of the focus as an ordered pair.

Graphy2=16xy^2=-16x. Give the equation of the directrix, as an equation.

Graphy2=16xy^2=-16x. Give both endpoints of the latus rectum as ordered pairs, separated by a comma.

Example. Graph x2=6yx^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 x2=4pyx^2=4py. Thus, the axis of symmetry is the yy-axis. It follows that:

  • 6=4p-6=4p, so p=32p=-\tfrac{3}{2}. Since p<0p<0, the parabola opens down.
  • the coordinates of the focus are (0,p)=(0,32)(0,p)=\left(0,-\tfrac{3}{2}\right)
  • the equation of the directrix is y=p=32y=-p=\tfrac{3}{2}
  • the endpoints of the latus rectum can be found by substituting y=32y=\tfrac{3}{2} into the original equation, (±3,32)\left(\pm3,-\tfrac{3}{2}\right)

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

Graphx2=8yx^2=8y. Give the coordinates of the focus as an ordered pair.

Graphx2=8yx^2=8y. Give the equation of the directrix, as an equation.

Graphx2=8yx^2=8y. Give both endpoints of the latus rectum as ordered pairs, separated by a comma.

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 xx- or yy-axis.
    • If the given coordinates of the focus have the form (p,0)(p,0), then the axis of symmetry is the xx-axis. Use the standard form y2=4pxy^2=4px.
    • If the given coordinates of the focus have the form (0,p)(0,p), then the axis of symmetry is the yy-axis. Use the standard form x2=4pyx^2=4py.
  2. Multiply 4p4p.
  3. Substitute the value from Step 2 into the equation determined in Step 1.

Example. What is the equation for the parabola with focus (12,0)\left(-\tfrac{1}{2},0\right) and directrix x=12x=\tfrac{1}{2}?

Solution. The focus has the form (p,0)(p,0), so the equation will have the form y2=4pxy^2=4px.

  • Multiplying 4p4p, we have 4p=4(12)=24p=4\left(-\tfrac{1}{2}\right)=-2.
  • Substituting for 4p4p, we have y2=4px=2xy^2=4px=-2x.

Therefore, the equation for the parabola is y2=2xy^2=-2x.

What is the equation for the parabola with focus(0,72)\left(0,\tfrac{7}{2}\right)and directrixy=72y=-\tfrac{7}{2}? Write the equation in standard form.

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 hh units horizontally and kk units vertically, the vertex will be (h,k)(h,k). This translation results in the standard form of the equation we saw previously with xx replaced by (xh)(x-h) and yy replaced by (yk)(y-k).

To graph parabolas with a vertex (h,k)(h,k) other than the origin, we use the standard form (yk)2=4p(xh)(y-k)^2=4p(x-h) for parabolas that have an axis of symmetry parallel to the xx-axis, and (xh)2=4p(yk)(x-h)^2=4p(y-k) for parabolas that have an axis of symmetry parallel to the yy-axis. These standard forms are given below, along with their general graphs and key features.

Standard Forms of Parabolas with Vertex (hh, kk). The table and figures below summarize the standard features of parabolas with a vertex at a point (h,k)(h,k).

Axis of SymmetryEquationFocusDirectrixEndpoints of Latus Rectum
y=ky=k(yk)2=4p(xh)(y-k)^2=4p(x-h)(h+p,k)(h+p,k)x=hpx=h-p(h+p,k±2p)(h+p,k\pm2p)
x=hx=h(xh)2=4p(yk)(x-h)^2=4p(y-k)(h,k+p)(h,k+p)y=kpy=k-p(h±2p,k+p)(h\pm2p,k+p)

(a) When p>0p>0, the parabola opens right.

(b) When p<0p<0, the parabola opens left.

(c) When p>0p>0, the parabola opens up.

(d) When p<0p<0, the parabola opens down.

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: (yk)2=4p(xh)(y-k)^2=4p(x-h) or (xh)2=4p(yk)(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 (yk)2=4p(xh)(y-k)^2=4p(x-h), then:
      • use the given equation to identify hh and kk for the vertex, (h,k)(h,k)
      • use the value of kk to determine the axis of symmetry, y=ky=k
      • set 4p4p equal to the coefficient of (xh)(x-h) in the given equation to solve for pp. If p>0p>0, the parabola opens right. If p<0p<0, the parabola opens left.
      • use h,k,h,k, and pp to find the coordinates of the focus, (h+p,k)(h+p,k)
      • use hh and pp to find the equation of the directrix, x=hpx=h-p
      • use h,k,h,k, and pp to find the endpoints of the latus rectum, (h+p,k±2p)(h+p,k\pm2p)
    • If the equation is in the form (xh)2=4p(yk)(x-h)^2=4p(y-k), then:
      • use the given equation to identify hh and kk for the vertex, (h,k)(h,k)
      • use the value of hh to determine the axis of symmetry, x=hx=h
      • set 4p4p equal to the coefficient of (yk)(y-k) in the given equation to solve for pp. If p>0p>0, the parabola opens up. If p<0p<0, the parabola opens down.
      • use h,k,h,k, and pp to find the coordinates of the focus, (h,k+p)(h,k+p)
      • use kk and pp to find the equation of the directrix, y=kpy=k-p
      • use h,k,h,k, and pp to find the endpoints of the latus rectum, (h±2p,k+p)(h\pm2p,k+p)
  3. Plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.

Example. Graph (y1)2=16(x+3)(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 (yk)2=4p(xh)(y-k)^2=4p(x-h). Thus, the axis of symmetry is parallel to the xx-axis. It follows that:

  • the vertex is (h,k)=(3,1)(h,k)=(-3,1)
  • the axis of symmetry is y=k=1y=k=1
  • 16=4p-16=4p, so p=4p=-4. Since p<0p<0, the parabola opens left.
  • the coordinates of the focus are (h+p,k)=(3+(4),1)=(7,1)(h+p,k)=(-3+(-4),1)=(-7,1)
  • the equation of the directrix is x=hp=3(4)=1x=h-p=-3-(-4)=1
  • the endpoints of the latus rectum are (h+p,k±2p)=(3+(4),1±2(4))(h+p,k\pm2p)=(-3+(-4),1\pm2(-4)), or (7,7)(-7,-7) and (7,9)(-7,9)

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

Graph(y+1)2=4(x8)(y+1)^2=4(x-8). Give the vertex as an ordered pair.

Graph(y+1)2=4(x8)(y+1)^2=4(x-8). Give the coordinates of the focus as an ordered pair.

Graph(y+1)2=4(x8)(y+1)^2=4(x-8). Give both endpoints of the latus rectum as ordered pairs, separated by a comma.

Example. Graph x28x28y208=0x^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 (xh)2=4p(yk)(x-h)^2=4p(y-k). Thus, the axis of symmetry is parallel to the yy-axis. To express the equation of the parabola in this form, we begin by isolating the terms that contain the variable xx in order to complete the square.

x28x28y208=0x28x=28y+208x28x+16=28y+208+16(x4)2=28y+224(x4)2=28(y+8)(x4)2=47(y+8) \begin{array}{lrcl} & 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\cdot7\cdot(y+8) \end{array}

It follows that:

  • the vertex is (h,k)=(4,8)(h,k)=(4,-8)
  • the axis of symmetry is x=h=4x=h=4
  • since p=7p=7, p>0p>0 and so the parabola opens up
  • the coordinates of the focus are (h,k+p)=(4,8+7)=(4,1)(h,k+p)=(4,-8+7)=(4,-1)
  • the equation of the directrix is y=kp=87=15y=k-p=-8-7=-15
  • the endpoints of the latus rectum are (h±2p,k+p)=(4±2(7),8+7)(h\pm2p,k+p)=(4\pm2(7),-8+7), or (10,1)(-10,-1) and (18,1)(18,-1)

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

Graph(x+2)2=20(y3)(x+2)^2=-20(y-3). Give the vertex as an ordered pair.

Graph(x+2)2=20(y3)(x+2)^2=-20(y-3). Give the equation of the directrix, as an equation.

Graph(x+2)2=20(y3)(x+2)^2=-20(y-3). Give both endpoints of the latus rectum as ordered pairs, separated by a comma.

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. 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. A cross-section of a design for a travel-sized solar fire starter is shown below. 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 the first part to find the depth of the fire starter.

Solution.

  • The vertex of the dish is the origin of the coordinate plane, so the parabola will take the standard form x2=4pyx^2=4py, where p>0p>0. The igniter, which is the focus, is 1.7 inches above the vertex of the dish. Thus we have p=1.7p=1.7.

    x2=4pyx2=4(1.7)yx2=6.8y \begin{array}{lrcl} & x^2 &=& 4py \\[4pt] & x^2 &=& 4(1.7)y \\[4pt] & x^2 &=& 6.8y \end{array}
  • The dish extends 4.52=2.25\tfrac{4.5}{2}=2.25 inches on either side of the origin. We can substitute 2.252.25 for xx in the equation from the first part to find the depth of the dish.

    x2=6.8y(2.25)2=6.8yy0.74 \begin{array}{lrcl} & x^2 &=& 6.8y \\[4pt] & (2.25)^2 &=& 6.8y \\[4pt] & y &\approx& 0.74 \end{array}

The dish is about 0.740.74 inches deep.

Balcony-sized solar cookers have been designed for families living in India. The top of a dish has a diameter of1,6001{,}600mm. The sun’s rays reflect off the parabolic mirror toward the “cooker,” which is placed320320mm 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 thexx-axis as its axis of symmetry). Write the equation in standard form.

Using the equation found for the solar cooker’s cross-section,y2=1280xy^2=1280x, find the depth of the cooker in millimeters.

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

Key equations

Parabola, vertex at origin, axis of symmetry on xx-axisy2=4pxy^2=4px
Parabola, vertex at origin, axis of symmetry on yy-axisx2=4pyx^2=4py
Parabola, vertex at (h,k)(h,k), axis of symmetry on xx-axis(yk)2=4p(xh)(y-k)^2=4p(x-h)
Parabola, vertex at (h,k)(h,k), axis of symmetry on yy-axis(xh)2=4p(yk)(x-h)^2=4p(y-k)

Key concepts

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

Practice

Graph parabolas with vertices at the origin

For the equationx=8y2x=8y^2, rewrite it in standard form.

For the equationx=8y2x=8y^2, rewritten in standard form asy2=18xy^2=\tfrac{1}{8}x, give the coordinates of the focus as an ordered pair.

For the equationy=4x2y=-4x^2, rewrite it in standard form.

For the equationy=4x2y=-4x^2, rewritten in standard form asx2=14yx^2=-\tfrac{1}{4}y, give the equation of the directrix, as an equation.

For the equationx=36y2x=36y^2, rewrite it in standard form.

For the equationx=36y2x=36y^2, rewritten in standard form asy2=136xy^2=\tfrac{1}{36}x, give both endpoints of the latus rectum as ordered pairs, separated by a comma.

Write equations of parabolas in standard form

The vertex of a parabola is(3,1)(-3,-1), and the endpoints of its latus rectum are(0,5)(0,5)and(0,7)(0,-7). Find the equation of the parabola, in standard form.

A parabola has vertex(0,0)(0,0), directrixy=4y=4, and focus(0,4)(0,-4). Find the equation of the parabola, in standard form.

The graph below shows a parabola with vertex(0,0)(0,0)and focus(0,14)\left(0,\tfrac{1}{4}\right), opening upward along a vertical axis of symmetry. Find the equation of the parabola, in standard form.

Graph parabolas with vertices not at the origin

Rewritey2+12x6y+21=0y^2+12x-6y+21=0in standard form by completing the square.

For the parabolay2+12x6y+21=0y^2+12x-6y+21=0, rewritten in standard form as(y3)2=12(x+1)(y-3)^2=-12(x+1), give the vertex as an ordered pair.

For the parabolay2+12x6y+21=0y^2+12x-6y+21=0, rewritten in standard form as(y3)2=12(x+1)(y-3)^2=-12(x+1), give the equation of the directrix, as an equation.

Rewritex24x+2y6=0x^2-4x+2y-6=0in standard form by completing the square, then give the coordinates of the focus as an ordered pair.

For the parabolax24x+2y6=0x^2-4x+2y-6=0, rewritten in standard form as(x2)2=2(y5)(x-2)^2=-2(y-5), give the axis of symmetry, as an equation.

Solve applied problems involving parabolas

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 asx2=4yx^2=4y. At what coordinates should you place the light bulb?

A satellite dish is shaped like a paraboloid of revolution, formed by rotating a parabola around its axis of symmetry, with the receiver at the focus. If the dish is1212feet across at its opening and44feet deep at its center, how many feet above the vertex should the receiver be placed?

An arch is in the shape of a parabola. It has a span of100100feet and a maximum height of2020feet. Find the equation of the parabola, with the vertex at the top of the arch and the ends of the span on thexx-axis, in standard form.

Using the arch equationx2=125(y20)x^2=-125(y-20), determine the height of the arch4040feet from the center, rounded to one decimal place.


This section is adapted from Precalculus 2e, Section 10.3: The Parabola by Jay Abramson and OpenStax, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: omitted a coreq-skills block the pinned CNXML prepends before the section proper (its own two-objective review of graphing vertical and horizontal parabolas from general or standard form, keyed to Intermediate Algebra 11.2.1/11.2.2, complete with its own worked examples and two “Practice Makes Perfect” exercise sets) — confirmed against the rendered PDF, page 1042 (true PDF index 1052), where Real-World Applications exercise 70 of the previous section runs directly into the “10.3 The Parabola” heading with no corequisite-skills material between them. Omitted the credit photograph of Katherine Johnson, keeping the two biographical paragraphs that introduce it. Recreated every instructional figure as an accessible spec-first SVG, each independently checked against the rendered PDF: the double-cone “Parabola” formation figure (Figure 2), drawn as an exact schematic — a right double cone and a cutting plane parallel to one generator, with the base circles, the cone silhouettes, and the parabolic trace all computed from the cone equation under a fixed oblique projection, hidden portions dashed — and corrected to match the section’s own prose (“If the plane is parallel to the edge of the cone, an unbounded curve is formed”) and the rendered art (a plane crossing only one nappe) rather than the pinned CNXML’s alt text for that figure — a source defect, detailed below; the generic key-features figure (Figure 3) and the point-on-the-parabola derivation figure (Figure 4), both drawn with representative numbers since the source states them symbolically; the four-panel standard-forms-at-the-origin figure (Figure 5) and its vertex-(h,k)(h,k) counterpart (Figure 9), each panel independently re-derived from p=±2p=\pm2 (and h=k=1h=k=1 for Figure 9) rather than traced from the source art, with the negative-pp panels’ latus-rectum labels kept in the source’s own 2p\lvert 2p\rvert form; the tangent-lines figure (Figure 6); both Example figures for the origin case (Figures 7 and 8) and both for the vertex-(h,k)(h,k) case (Figures 10 and 11); the reflecting-property schematic (Figure 12), with three representative parallel rays reflecting to the focus (the source’s exact ray count and angles are not stated); and the fire-starter cross-section (Figure 13), the given diagram for that example, drawn to the stated 1.71.7-inch focus height and 4.54.5-inch width with its own (shorter, unlabeled) depth marked separately, matching the source’s convention of leaving the solved depth off the given figure. Kept the “Media” callout’s introductory sentence but omitted its four external video links, matching house precedent elsewhere in this book. Every “write in standard form” fill-in declares answerForm="parabola-standard-form" (added to the grading engine for this chapter), replayed against its own general-form span to confirm the token reports form rather than correct; focus and vertex are keyed as ordered pairs, directrix and axis of symmetry as equations (“as an equation” stated in the question, since a bare number also grades correct against the equation but the reverse does not hold), and latus rectum endpoints as an answerMode="unordered" two-member list, per this book’s established convention for a \pm-unparseable pair. Two end-of-section “given vertex, focus, and directrix” items (exercises 47 and 49 in the Graphical set) were independently solved but not used: both have an irrational 4p4p (a 2\sqrt2 or 5\sqrt5 coefficient on the linear term), and parabola-standard-form’s coefficient pattern accepts only an integer, decimal, or written fraction — the correct keyed answer for either item reports form against itself, so neither is retype-safe under the token this chapter requires. The three algebraic “rewrite in standard form, then determine the vertex, focus, and directrix” exercises used for the vertex-at-the-origin objective keep only the standard-form rewrite and one non-trivial feature (focus, directrix, or latus rectum) as components, omitting a redundant “give the vertex” component, since the vertex is (0,0)(0,0) for every exercise in that family by construction and restating it tests nothing the standard-form component does not already require. “Determine the equation for the parabola from its graph” exercise 51 (vertex (0,0)(0,0), focus (0,14)\left(0,\tfrac14\right), both printed directly on the source’s own graph) was recreated as an accessible given-figure feeding a standard-form fill-in, the same treatment the fire-starter’s given diagram receives; two sibling items in that set (exercises 52, 54) carry no printed solution and were not used. Eleven selected end-of-section exercises were adapted into eighteen interactive Practice components across four objective groups, every one independently re-derived (including by running the arithmetic in Node) rather than read off the source key.