Question

Derive the equation of the parabola with a focus at (-7, 5) and a directrix of y = -11.

Question 2
Derive the equation of the parabola with a focus at (-5, 5) and a directrix of y = -1

Derive the equation of the parabola with a focus at (0, 1) and a directrix of y = -1

Answer 1

ok so, this is not something that i do often but here goes. do you know the generic equation for a parabola?

Answer 2

I have to go now so i will leave you with the equation and an example.
Generic equation for a parabola:\[(x-h)^{2} = 4p(y-k)\]
the coordinates of the vertex are:
\[Vertex - (h,k)\]
The Coordinates of the vertex are:\[Focus - (h,(k+p))\]
The equation for the directix:
\[y = k-p\]

Therefore your first question - Focus = (-7,5), Directix = -11
From the equation for a directix:
\[-11=y=k-p\]
From the y coordinate for the focus:\[5 = k-p\]
You can then solve these equations simultaneously and get k and p:
k = -3, p = 8 you know from the x coodinate of the focus is h = -7 so you can wright the equation for the parabola\[(x-(-7))^{2}=4\times8(y-(-3))\]Simplifies to :\[(x+7)^{2} = 32(y+3)\]