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Parabola Help (page 3)

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By — McGraw-Hill Professional
Updated on Oct 4, 2011

Finding the Equation of a Parabola

If we know any two of the vertex, focus, and directrix, we can find an equation of the parabola. From the information given, we first need to decide which form to use. Knowing the directrix is the fastest way to decide this. If the directrix is a horizontal line ( y = number), then the equation is ( xh ) 2 = 4 p ( yk ). If the directrix is a vertical line ( x = number), then the equation is ( yk ) 2 = 4 p ( xh ). If we do not have the directrix, we need to look at the coordinates of the vertex and focus. Either both the x -coordinates will be the same or both y -coordinates will be. If both x -coordinates are the same, the parabola opens up or down. We need to use the form ( xh ) 2 = 4 p ( yk ). If both y -coordinates are the same, the parabola opens to the side. We need to use the form ( yk ) 2 = 4 p ( xh ). Once we have decided which form to use, we might need to use algebra to find h , k , and p . For example, if we know the focus is (2, −1) and the directrix is x = 6, then we know hp = 6 and h + p = 2 and k = −1. The equations hp = 6 and h + p = 2 form a system of equations.

Conic Sections Parabolas

Conic Sections Parabolas

Now that we have all three numbers and the form, we are ready to write the equation: ( y + 1) 2 = −8( x − 4).

Examples

Find an equation for the parabola.

  • The directrix is y = 2, and the vertex is (3, 1).
  • Because the directrix is a horizontal line, the equation we want is ( xh ) 2 = 4 p ( yk ). The vertex is (3, 1), giving us h = 3 and k = 1. From y = kp and y = 2, we have 1 − p = 2, making p = −1. The equation is ( x − 3) 2 = −4( y − 1).

  • The focus is (4, −2), and the vertex is (0, −2).
  • The y -coordinates are the same, so this parabola opens to the side, and the equation we need is ( yk ) 2 = 4 p ( xh ). The vertex is ( h , k ) = (0, −2), giving us h = 0 and k = −2. The focus is ( h + p , k ) = (4, −2). From this we have h + p = 0 + p = 4, making p = 4. The equation is ( y + 2) 2 = 16 x .

Conic Sections Parabolas

Fig. 12.7

The directrix is the vertical line x = −1, and the focus is (3, 2). Because the parabola opens to the right, the form we need is ( yk ) 2 = 4 p ( xh ).

From the focus we have ( h + p , k ) = (3, 2), so h + p = 3 and k = 2. The directrix is x = −1 and x = hp so hp = −1.

Conic Sections Parabolas

Parabola Practice Problems

Practice

  1. Identify the vertex, focus, and directrix for ( y −5) 2 = 10( x − 1).
  2. Identify the vertex, focus, and directrix for Conic Sections Parabolas .
  3. Find an equation for the parabola that has directrix y = − 2 and focus (4, 10).

For Problems 4-6, match the equation with its graph in Figures 12.8-12.10.

Conic Sections Parabolas

Fig. 12.8

Conic Sections Parabolas

Fig. 12.9

Conic Sections Parabolas

Fig. 12.10

  1. x 2 = −8( y + 1)
  2. ( x + 1) 2 = 4( y − 2)
  3. ( y − 2) 2 = −6( x + 3)

Solutions

  1. h = 1, k = 5, and Conic Sections Parabolas (from 4 p = 10). The vertex is (1, 5); the focus is Conic Sections Parabolas and the directrix is Conic Sections Parabolas (from Conic Sections Parabolas.

  2. h = −6, k = 4, and Conic Sections Parabolas . The vertex is (−6, 4); the focus is Conic Sections Parabolas ; and the directrix is Conic Sections Parabolas.

  3. We want to use the equation ( xh ) 2 = 4 p ( yk ). The focus is ( h , k + p ), so h = 4 and k + p = 10. The directrix is y = kp , so kp = −2.

    Conic Sections Parabolas

    4 + p = 10 Let k = 4 in k + p = 10

    p = 6

    The equation is ( x - 4) 2 = 24( y − 4).

  4. Figure 12.9

  5. Figure 12.8

  6. Figure 12.10

Practice problems for this concept can be found at: Conic Sections Practice Test.

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