Find the equation in standard form of the parabola with focus (-2,-3) and directrix y = 3?

Question

I have a midterm tomorrow and pretty much understand every question in the review packet except this one!

I know the answer as we have a key but I don't see how she got there.

Thank you!

Answer 1

Find the equation in standard form of the parabola with focus (-2,-3) and directrix y = 3.

The directrix is a horizontal line so the parabola is vertical.
The focus is below the directrix so the parabola opens downward.

First get the equation of the parabola in vertex form and then convert it to standard form. The parabola with vertex (h,k) has the form:

4p(y - k) = (x - h)²
where p is the directed distance from the vertex to the focus.

The directed distance from the directrix to the focus is 2p.

2p = -3 - 3 = -6
p = -3
4p = -12

The vertex is the midpoint between the directrix and the focus.

(h, k) = (-2, 0).

The equation of the parabola is:

4p(y - k) = (x - h)²
-12(y - 0) = (x + 2)²
-12y = x² + 4x + 4
y = -x²/12 - x/3 - 1/3

Answer 2

The distance between the directrix and the focus is halved by the vertex. The distance from a point on the parabola to the directrix, must be the same as its distance from the focus. (X - 3)^2 + Y^2 = (X + 3)^2 x^2 + 6x + 9 - x^2 + 6x - 9 = y^2, and you get your expression up there: y^2 = 12x

Answer 3

This site will help you:

http://www.personal.kent.edu/~rmuhamma/Algorithms/MyAlgorithms/parabola.htm

Guido

Answer 4

is that to do with graphs?