Question

Can somebody tech me math derivatives please?
1. 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

o do these questions, you should find the vertex.
the vertex is exactly between the focus and the directrix.

The first step is plot the focus and the directrix, just to see what is what.
then find the point mid way between them.

can you do that ?

Answer 2

you probably want to use the form
y= a(x-h)^2 + k
where (h,k) is the vertex. (notice that once you have the vertex you have 2 out of the 3 numbers you need in this equation. So finding the vertex is very helpful.

Answer 3

Also, you will need the distance between the vertex and the focus (which is ½ the whole distance between the focus and the directrix)

call the distance between the vertex and the focus P
we will need P to finish the problem.

Answer 4

Meanwhile, maybe Khan can help
http://www.khanacademy.org/math/trigonometry/conics_precalc/parabolas_precalc/v/parabola-focus-and-directrix-1

Answer 5

okay one second... This is everything I've done, up to the point I'm stuck at.
for the focus and directrix i use the distance formula \[D=\sqrt{(X _{1}-X _{2})^{2}}+(Y _{2}-Y _{1})^{2}\]
ALL OF IT SHOULD BE IN THE SQUARE ROOT SIGN.
then I substituted the focus coordinates for \[(X _{1},Y _{1}) \]
then i put the directrix Y=-15 into the point \[(X _{2},Y _{2})\]
and got \[\sqrt{(X _{2}-4)^2+(Y _{2}-(-7)^2}=\sqrt{(Y _{2}+15})^2\]
then square root everything

Answer 6

I got
\[f(x)=\frac{ 1 }{ 16 }x ^{2}-\frac{ 1 }{ 2 }x-10\]
Is that right?

Answer 7

yes, that is very impressive.

Answer 8

You used the definition of a parabola to find the equation.

I am guessing that class is teaching a different method, which you might have to learn ?

Answer 9

I'm not sure what its teaching but its difficult.

Answer 10

another way to do it is:
focus is (4,-7) and directrix is y= -15

it looks like this (roughly):