Parabolas Not Centered 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 2 and Figure 9 summarize the standard features of parabolas with a vertex at a point $(h,k)$.

Table 2

Figure 9 (a) When $p > 0$, the parabola opens right. (b) When $p < 0$, the parabola opens left. (c) When $p > 0$, the parabola opens up. (d) When $p < 0$, 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: $(y−k)^2=4p(x−h)$ 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.
   1. 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 $h$ and $p$ to find the equation of the directrix, $x=h−p$
      • use $h,k$, and $p$ to find the endpoints of the latus rectum, $(h+p,k±2p)$
   2. 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, $(h±2p,k+p)$
3. Plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.

EXAMPLE 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 $(h+p,k±2p)=(−3+(−4),1±2(−4))$, 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. See Figure 10.

Figure 10

TRY IT #4

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

EXAMPLE 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.

$\begin{aligned} x^{2}-8 x-28 y-208 &=0 \\ x^{2}-8 x &=28 y+208 \\ x^{2}-8 x+16 &=28 y+208+16 \\ (x-4)^{2} &=28 y+224 \\ (x-4)^{2} &=28(y+8) \\ (x-4)^{2} &=4 \cdot 7 \cdot(y+8) \end{aligned}$

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 $(h±2p,k+p)=(4±2(7),−8+7)$, 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. See Figure 11.

Figure 11

TRY IT #5

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

Source: Rice University, https://openstax.org/books/college-algebra/pages/8-3-the-parabola
This work is licensed under a Creative Commons Attribution 4.0 License.

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 12. 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.

EXAMPLE 6

Solving Applied Problems Involving Parabolas

A cross-section of a design for a travel-sized solar fire starter is shown in Figure 13. 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 ⓐ to find the depth of the fire starter.

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

Solution

• ⓐ 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 $p=1.7$.

$\begin{array}{ll} x^{2}=4 p y & \text { Standard form of upward-facing parabola with vertex }(0,0) \\ x^{2}=4(1.7) y & \text { Substitute } 1.7 \text { for } p . \\ x^{2}=6.8 y & \text { Multiply. } \end{array}$

• ⓑ The dish extends $\frac{4.5}{2}=2.25$ 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.

$\begin{array}{ll} x^2&=6.8y & \text { Equation found in part (a). } \\ (2.25)^2&=6.8y & \text { Substitute } 2.25 \text { for } x . \\ y &\approx 0.74 & \text { Solve for } y. \end{array}$

TRY IT #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.

ⓐ 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 ⓐ to find the depth of the cooker.