Question

Derive the equation of the parabola with a focus at (−2, 4) and a directrix of y = 6. Put the equation in standard form

Answer 1

If I consider a point P=(x,y) of your parabola, then applying the definition of a parabola, I can write this:
\[(y-4)^{2}+(x+2)^{2}=(y-6)^{2}\]
please develop this equality

Answer 2

So the vertex would be y=5 ..... and i think x=-2
Though standard form is \[f(x)=a(x-h)^2+k\] I still need (a) which i think is 1?

Answer 3

@campbell_st I need your help dude

Answer 4

from the equation above, I got:
\[a=-\frac{ 1 }{ 4 }\]

Answer 5

... Are you sure?

Answer 6

yes!

Answer 7

I can give you the equation of your parabola, namely:
\[y=-\frac{ 1 }{ 4 }(x+2)^{2}+5\]
is it that, please?

Answer 8

Why do you have two y values in the equation when I'm supposed to be using standard form? I'm confused how a is 1/4

Answer 9

I just don't understand the 1/4 out of all that

Answer 10

please note that I said that a=-1/4

Answer 11

Oh, sorry, a=-1/4
Could you explain to me how you got that answer?

Answer 12

please as I said before, please develop the above expression that I wrote in my first post

Answer 13

But how did you develop the expression?

Answer 14

for example:
\[(y-4)^{2}=y ^{2}+16-8y\]
now, please do the same with the remaining parentheses

Answer 15

... Oh