Question

Which is the standard form of the equation of the parabola that has a vertex of (–4, –6) and a directrix of y = 3?

Answer 1

Hello,

focus=(-4,-15)

If focus =(a,b) and directrix y=k
the parabola is:
y=1/(2(b-k)) * (x-a)²+(b+k)/2

So y=1/(2*(-15-3))*(x+4)²+(-15+3)/2
y=-(x+4)²/36 -6

Answer 2

The required equation is
`(x+4)^2=-36(y+6)`.

Explanation:

The standard form of the equation of the parabola is

`(x-h)^2=4p(y-k)` .... (1)

where, (h, k) is vertex and y = k - p is directrix.

It is given that vertex of parabola is (–4, –6) and the directrix is y = 3.

`(-4,-6)=(h,k)`

On comparing both the sides, we get

`h=-4,k=-6`

Directrix of the parabola is

`y=k-p`

Put y=3 and k=-6 in the above equation.

`3=-6-p`

`3+6=-p`

`9=-p`

`-9=p`

Substitute h=-4,p=-9 and k=-6 in equation (1).

`(x-(-4))^2=4(-9)(y-(-6))`

`(x+4)^2=-36(y+6)`

Therefore the required equation is
`(x+4)^2=-36(y+6)`.