Derive the equation of the parabola with a focus at (-7,5) and a directrix of y=-11.
f(x)=1/32(x+7)^2-3
f(x)=-1/32(x+7)^2-3
f(x)=-1/32(x-7)^2-3
f(x)=1/32(x-7)^2-3

Derive the equation of the parabola with a focus at (2,-1) and a directrix of y=-1/2
(x)=-(x+2)^2-3/4
f(x)=(x+2)^2+3/4
f(x)=-(x-2)^2+3/4
f(x)=(x-2)^2-3/4

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

Question

Derive the equation of the parabola with a focus at (-7,5) and a directrix of y=-11. 
f(x)=1/32(x+7)^2-3 
f(x)=-1/32(x+7)^2-3 
f(x)=-1/32(x-7)^2-3 
f(x)=1/32(x-7)^2-3 

Derive the equation of the parabola with a focus at (2,-1) and a directrix of y=-1/2 
(x)=-(x+2)^2-3/4 
f(x)=(x+2)^2+3/4 
f(x)=-(x-2)^2+3/4 
f(x)=(x-2)^2-3/4 

derive the equation of the parabola with a focus at (-2 4) and a directrix of y=6. Put the equation in standard form.

savannah_noelle · Answer

@mathstudent55 @math&ing001 @MathKat can you guys help please?

jdoe0001 · Answer

http://www.youtube.com/watch?v=XwtyQsCLnEE