Question

What is the equation of the parabola with a focus at (-1,2) and a directrix of `y=1/2`?

Answer 1

The equation of the parabola is `(x+1)^2=3(y-5/4)`

Explanation:

Any point `(x,y)` on the parabola is equidistant from the directrix and the focus.

Therefore,

`y-(1/2)=sqrt((x-(-1))^2+(y-(2))^2)`

`y-1/2=sqrt((x+1)^2+(y-2)^2)`

Squaring and developing the `(y-2)^2` term and the LHS

`(y-1/2)^2=(x+1)^2+(y-2)^2`

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

`(x+1)^2=3y-15/4=3(y-5/4)`

The equation of the parabola is `(x+1)^2=3(y-5/4)`

graph{((x+1)^2-3(y-5/4))(y-1/2)((x+1)^2+(y-2)^2-0.04)=0 [-12.66, 12.65, -6.33, 6.33]}