Question

Derive the equation of the parabola with a focus at (4, −7) and a directrix of y = −15. Put the equation in standard form.

Answer 1

@aaronq @Alchemista @SithsAndGiggles @sourwing @dan815 @freckles @ganeshie8 @goatdude101 @hhelpplzzzz @KamiBug @LazyBoy @zepdrix @cwrw238 @mathstudent55 @Mimi_x3 @mathmath333 Please help ;-;

Answer 2

Hmm, so the distance between the point and the line are always equal

Answer 3

I mean the point and parabola and the directrix and parapola.

Answer 4

f(x) = (1/12)x^2 - 8x + 11
f(x) = (1/16)x^2 - 8x - 10
f(x) = (1/16)x2 - (1/2)x + 11
f(x) = (1/16)x^2 - (1/2)x - 10

Answer 5

wio can you check my answer after this?

Answer 6

\[
\sqrt{(y-(-15))^2} = \sqrt{(x-4)^2+(y-(-7))^2}
\]Square both sides: \[
(y+15)^2=(x-4)^2+(y+7)^2
\]Simplify:\[
y^2+30y+225=x^2-8x+16 +y^2+14y+49\\
16y = x^2-8x-160\\
y = \frac{x^2-8x-160}{16}
\]

Answer 7

I used a long approach, there is another approach involving \[
4py= x^2
\]Though I don't completely remember it.

Answer 8

So its D?

Answer 9

Well, let \(x=4\). See what the value of \(y\) is. If the distances check out, you have your solution

Answer 10

please check wio

Answer 11

I got -26...