Parabola Help

Match the graphs in Figures 12.3 through 12.6 with their equations.

Fig. 12.3

Fig. 12.4

Fig. 12.5

Fig. 12.6

• x ² = 6 y

• The equation is in the form ( x − h ) ² = 4 p ( y − k ), so the parabola will open up or down. We have . Now we know three things—that the parabola opens up (because p is positive), that the focus is ), and the directrix is (from The graph that behaves this way is in Figure 12.5 .

• 

  Fig. 12.5

• y ² = 6 x

• The equation is in the form ( y − k ) ² = 4 p ( x − h ), so the parabola opens to the left or to the right. We have . Now we know that the parabola opens to the right, that the focus is , and that the directrix is . The graph for this equation is in Figure 12.3 .

• 

  Fig. 12.3

• y ² = –6 x

• The equation is in the form ( y − k ) ² = 4 p ( x − h ), so the parabola opens to the left or to the right. We have . The parabola opens to the left, the focus is , and the directrix is . The graph for this equation is in Figure 12.4.

• 

  Fig. 12.4

• x ² = –6 y

• The equation is in the form ( x − h ) ² = 4 p ( y − k ), so the parabola opens up or down. We have . The parabola opens down, the focus is . The directrix is (from . The graph for this equation is in Figure 12.6.

• 

  Fig. 12.6

Find the vertex, focus, and directrix. Determine if the parabola opens up, down, to the left, or to the right.

• ( x − 3) ² = 4( y − 2)

• This equation is in the form ( x − h ) ² = 4 p ( y − k ). The vertex is (3, 2). Once we have found p , we can find the focus and directrix and how the parabola opens. p = 1 (from 4 = 4 p ). The parabola opens up because p is positive; the focus is ( h , k + p ) = (3, 2 + 1) = (3, 3); and the directrix is y = 1 (from y = k − p = 2 − 1 = 1).

• ( y + 1) ² = 8( x − 3)

• The equation is in the form ( y − k ) ² = 4 p ( x − h ). The vertex is (3, −1), p = 2 (from 8 = 4 p ); the parabola opens to the right; the focus is ( h + p , k ) = (3 + 2, −1) = (5, −1); and the directrix is x = 1 (from x = h − p = 3−2 = 1).

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 ( x − h ) ² = 4 p ( y − k ). The vertex is (3, 1), giving us h = 3 and k = 1. From y = k − p 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 ( y − k ) ² = 4 p ( x − h ). 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 .

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 ( y − k ) ² = 4 p ( x − h ).

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