Question

Need help!!

Derive the equation of the parabola with a focus at (2, −1) and a directrix of y = −one half.

Answer 1

F(x)= -.5(x+2)-1 this should be your equation because the .5 is your scale factor, +2 will move you to the right, and -1 will shift your point down one.

Answer 2

The equation of required parabola is
`y=-(x-2)^2-(3)/(4)`.

Explanation:

The standard form of a parabola is

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

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

It is given that the focus of the parabola is at (2,-1).

`(h,k+p)=(2,-1)`

`h=2`

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

The directrix of the parabola is

`y=-(1)/(2)`

`k-p=-(1)/(2)` .... (2)

Add equation (1) and (2).

`2k=-(3)/(2)`

Divide both sides by 2.

`k=-(3)/(4)`

Put this value in equation (1).

`-(3)/(4)+p=-1`

`p=-1+(3)/(4)`

`p=-(1)/(4)`

Substitute h=2,
`k=-(3)/(4)` and
`p=-(1)/(4)` in the standard form of the parabola.

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

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

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

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

Therefore the equation of required parabola is
`y=-(x-2)^2-(3)/(4)`.