Question

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

Answer 1

`y=1/8(x^2-2x-7)`

Explanation:

Parabola is the locus of a point `(x,y)` which moves so thatits distance from a point called focus, here `(1,1)`, and a line called directrix `y=-3` or `y+3=0` is always equal.

Here distance of `(x,y)` from `(1,1)` is

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

and distance of `(x,y)` from `y+3=0` is `|y+3|`

Hence equation of parabola is `sqrt((x-1)^2+(y-1)^2)=|y+3|` and on squaring it becomes

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

or `x^2-2x+1+y^2-2y+1=y^2+6y+9`

or `x^2-2x-7=8y`

or `y=1/8(x^2-2x-7)`

in vertex form this becomes `y=1/8(x-1)^2-1`

graph{(x^2-2x-7-8y)(y+3)((x-1)^2+(y-1)^2-0.02)=0 [-10, 10, -5, 5]}