Question

HELP!!! Using a directrix of y = 2 and a focus of (3, −4), what quadratic function is created?

Answer 1

parabola with vertx (3,-1) that opens down

4(-3)(y+1)=(x-3)^2
y+1=(-1/12)(x-3)^2
y=(-1/12)(x-3)^2-1

Answer 2

The required quadratic function is
`y=-(1)/(12)(x-3)^2-1`.

Explanation:

The standard form of the parabola is

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

Where the focus is (h, k + p) and the directrix is y = k - p.

The directrix of y = 2 and a focus of (3, −4).

`(h,k+p)=(3,-4)`

On comparing both sides we get

`h=3`

`k+p=-4` ...... (1)

`y=k-p`

`k-p=2` ...... (2)

Add equation (1) and (2).

`2k=-2`

`k=-1`

Substitute k=-1 in equation (1).

`(-1)+p=-4`

`p=-3`

Therefore the equation of parabola is

`(x-3)^2=4(-3)(y-(-1))`

`(x-3)^2=-12(y+1)`

It can be rewritten as

`y=-(1)/(12)(x-3)^2-1`

Therefore the required quadratic function is
`y=-(1)/(12)(x-3)^2-1`.