What is the equation in standard form of the parabola with a focus at (3,6) and a directrix of y= 7?

Question

1412 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-29T20:48:32

The equation is $y = - \frac{1}{2} {(x - 3)}^{2} + \frac{13}{2}$ $y=-1/2(x-3)^2+13/2$

A point on the parabola is equidistant from the directrix and the focus.

The focus is $F = (3, 6)$ $F=(3,6)$

The directrix is $y = 7$ $y=7$

$\sqrt{{(x - 3)}^{2} + {(y - 6)}^{2}} = 7 - y$ $sqrt((x-3)^2+(y-6)^2)=7-y$

Squaring both sides

${(\sqrt{{(x - 3)}^{2} + {(y - 6)}^{2}})}^{2} = {(7 - y)}^{2}$ $(sqrt((x-3)^2+(y-6)^2))^2=(7-y)^2$

${(x - 3)}^{2} + {(y - 6)}^{2} = {(7 - y)}^{2}$ $(x-3)^2+(y-6)^2=(7-y)^2$

${(x - 3)}^{2} + y^{2} - 12 y + 36 = 49 - 14 y + y^{2}$ $(x-3)^2+y^2-12y+36=49-14y+y^2$

$14 y - 12 y - 49 = {(x - 3)}^{2}$ $14y-12y-49=(x-3)^2$

$2 y = - {(x - 3)}^{2} + 13$ $2y=-(x-3)^2+13$

$y = - \frac{1}{2} {(x - 3)}^{2} + \frac{13}{2}$ $y=-1/2(x-3)^2+13/2$

graph{((x-3)^2+2y-13)(y-7)((x-3)^2+(y-6)^2-0.01)=0 [-2.31, 8.79, 3.47, 9.02]}