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Guias de estudo > College Algebra

Parabolas with Vertices Not at the Origin

Learning Objectives

  • Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the focal diameter given the equation of a parabola in standard form
  • Find the equation of a parabolic shaped object given dimensions
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)\left(h,k\right). This translation results in the standard form of the equation we saw previously with xx replaced by (x−h)\left(x-h\right) and yy replaced by (y−k)\left(y-k\right). To graph parabolas with a vertex (h,k)\left(h,k\right) other than the origin, we use the standard form (y−k)2=4p(x−h){\left(y-k\right)}^{2}=4p\left(x-h\right) for parabolas that have an axis of symmetry parallel to the x-axis, and (x−h)2=4p(y−k){\left(x-h\right)}^{2}=4p\left(y-k\right) 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.

A General Note: Standard Forms of Parabolas with Vertex (h, k)

The table summarizes the standard features of parabolas with a vertex at a point (h,k)\left(h,k\right).
Axis of Symmetry Equation Focus Directrix Endpoints of focal diameter
y=ky=k (y−k)2=4p(x−h){\left(y-k\right)}^{2}=4p\left(x-h\right) (h+p, k)\left(h+p,\text{ }k\right) x=h−px=h-p (h+p, k±2p)\left(h+p,\text{ }k\pm 2p\right)
x=hx=h (x−h)2=4p(y−k){\left(x-h\right)}^{2}=4p\left(y-k\right) (h, k+p)\left(h,\text{ }k+p\right) y=k−py=k-p (h±2p, k+p)\left(h\pm 2p,\text{ }k+p\right)
(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: (y−k)2=4p(x−h){\left(y-k\right)}^{2}=4p\left(x-h\right) or (x−h)2=4p(y−k){\left(x-h\right)}^{2}=4p\left(y-k\right).
  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 focal diameter.
    1. If the equation is in the form (y−k)2=4p(x−h){\left(y-k\right)}^{2}=4p\left(x-h\right), then:
      • use the given equation to identify hh and kk for the vertex, (h,k)\left(h,k\right)
      • use the value of kk to determine the axis of symmetry, y=ky=k
      • set 4p4p equal to the coefficient of (x−h)\left(x-h\right) 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,kh,k, and pp to find the coordinates of the focus, (h+p, k)\left(h+p,\text{ }k\right)
      • use hh and pp to find the equation of the directrix, x=h−px=h-p
      • use h,kh,k, and pp to find the endpoints of the focal diameter, (h+p,k±2p)\left(h+p,k\pm 2p\right)
    2. If the equation is in the form (x−h)2=4p(y−k){\left(x-h\right)}^{2}=4p\left(y-k\right), then:
      • use the given equation to identify hh and kk for the vertex, (h,k)\left(h,k\right)
      • use the value of hh to determine the axis of symmetry, x=hx=h
      • set 4p4p equal to the coefficient of (y−k)\left(y-k\right) 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,kh,k, and pp to find the coordinates of the focus, (h, k+p)\left(h,\text{ }k+p\right)
      • use kk and pp to find the equation of the directrix, y=k−py=k-p
      • use h,kh,k, and pp to find the endpoints of the focal diameter, (h±2p, k+p)\left(h\pm 2p,\text{ }k+p\right)
  3. Plot the vertex, axis of symmetry, focus, directrix, and focal diameter, and draw a smooth curve to form the parabola.

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

Graph (y−1)2=−16(x+3){\left(y - 1\right)}^{2}=-16\left(x+3\right). Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the focal diameter.

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

  • the vertex is (h,k)=(−3,1)\left(h,k\right)=\left(-3,1\right)
  • 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)\left(h+p,k\right)=\left(-3+\left(-4\right),1\right)=\left(-7,1\right)
  • the equation of the directrix is x=h−p=−3−(−4)=1x=h-p=-3-\left(-4\right)=1
  • the endpoints of the focal diameter are (h+p,k±2p)=(−3+(−4),1±2(−4))\left(h+p,k\pm 2p\right)=\left(-3+\left(-4\right),1\pm 2\left(-4\right)\right), or (−7,−7)\left(-7,-7\right) and (−7,9)\left(-7,9\right)
Next we plot the vertex, axis of symmetry, focus, directrix, and focal diameter, and draw a smooth curve to form the parabola.

Try It

Graph (y+1)2=4(x−8){\left(y+1\right)}^{2}=4\left(x - 8\right). Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the focal diameter.

Answer: Vertex: (8,−1)\left(8,-1\right); Axis of symmetry: y=−1y=-1; Focus: (9,−1)\left(9,-1\right); Directrix: x=7x=7; Endpoints of the latus rectum: (9,−3)\left(9,-3\right) and (9,1)\left(9,1\right).

Try It

In the graph below you will find a parabola whose vertex is (h,k)(h,k). The equation used to generate the parabola is (y−k)2=4p(x−h)(y-k)^2 = 4p(x-h). Your task in this exercise is to plot the vertex, axis of symmetry, focus, directrix, and endpoints of the focal diameter in terms of the free variables y,k,py,k,p.   For example, to plot the vertex, you would create a point (h,k)(h,k). https://www.desmos.com/calculator/8bj8n72isi

Answer:  

https://www.desmos.com/calculator/gvpu51ti7n
 

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.
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: Solving Applied Problems Involving Parabolas

A cross-section of a design for a travel-sized solar fire starter. 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.
Cross-section of a travel-sized solar fire starter

Answer:

  1. The vertex of the dish is the origin of the coordinate plane, so the parabola will take the standard form x2=4py{x}^{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=4pyStandard form of upward-facing parabola with vertex (0,0)x2=4(1.7)ySubstitute 1.7 for p.x2=6.8yMultiply.\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}
  2. The dish extends 4.52=2.25\frac{4.5}{2}=2.25 inches on either side of the origin. We can substitute 2.25 for xx in the equation from part (a) to find the depth of the dish.  x2=6.8yEquation found in part (a).(2.25)2=6.8ySubstitute 2.25 for x. y≈0.74Solve for y.\begin{array}{ll}\text{ }{x}^{2}=6.8y\hfill & \text{Equation found in part (a)}.\hfill \\ {\left(2.25\right)}^{2}=6.8y\hfill & \text{Substitute 2}\text{.25 for }x.\hfill \\ \text{ }y\approx 0.74 \hfill & \text{Solve for }y.\hfill \end{array} The dish is about 0.74 inches deep.

Try It

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.

Answer:

  1. y2=1280x{y}^{2}=1280x
  2. The depth of the cooker is 500 mm

 

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