Question

Find the equation of parabola whose focus is `(-1,2)` and directrix is `x=-3`?

Answer 1

`x=1/4y^2-y-1`

Explanation:

Parabola is the locus of a point which moves so that its distance from a point called focus and a line called directrix is always constant.

Let this point be `(x,y)`. Its distance from focus `(-1,2)` is

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

and its distance from `x=-3` or `x+3=0` is

`|x+3|`

As the two are equal we have

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

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

i.e. `x^2+2x+1+y^2-4y+4=x^2+6x+9`

or `-4x=-y^2+4y+4`

or `x=1/4y^2-y-1`

graph{(4x-y^2+4y+4)((x+1)^2+(y-2)^2-0.02)(x+3)=0 [-10, 10, -2.8, 7.2]}