Find the standard form of the equation of the parabola with a focus at (-4, 0) and a directrix at x = 4.

Question

So basically, it's a multiple choice question and I got stuck. So there are 4 options, 2 of them are completley wrong but the other two options are these: y=(-1/16)x^2, x=(-1/16)x^2

 

This is the formula I followed in order to solve this problem: x^2=4py   because this is what's written in my school textbook.

This is how I solved this problem:

x^2=4(-4)y

x^2=-16y

y=-1/16(x^2)

 

Did I solve this correctly? I'm not sure because the answer is either this: y=(-1/16)x^2 or this: x=(-1/16)x^2

So whuch one is it and can you please explain why????? 

Also, it says that the directrix is postive 4? Isn't directrix always suppose to be negative and the "P" is suppose to be positive?????? 

 

Answer 1

The directrix and focus will be opposite signs on a graph with vertex at (0,0). So if the focus is (-4,0)  the directrix will be at x=4.

 

Also the vertex , focus, and directrix will always be on the same line.

Answer 2

The answer should be y² = -16x

 

The vertex (h,k) of the parabola is the midpoint between

the focus and the directrix...

that means the vertex is at (0,0) for the parabola.

 

the parabola opens away from the directrix and towards

the focus so it opens left and the squared term is y instead of x

 

The magnitude of p is the absolute value of the

distance from the vertex to the focus and/or the

distance from the vertex to the directrix...

Absolute value of p = 4

 

Since the parabola opens away from the directrix

it opens left and p will be negative so p = -4

 

Your equation is  (y-k)² = 4p(x-h)

(y-0)² = 4(-4)(x-0)

y² = -16x