Question

How to find the equation of a parabola whose focus is `(1,1)` and directrix is `x-y-3=0`?

Answer 1

Equation of parabola is `x^2+2xy+y^2+2x-10y-5=0`

Explanation:

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

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

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

and its distance from a given line `x-y-3=0` is

`|(x-y-3)/(sqrt(1^2+1^2))|=|(x-y-3)/sqrt2|`

Hence equation of parabola is

`sqrt((x-1)^2+(y-1)^2)=|(x-y-3)/sqrt2|` and squaring

`(x-1)^2+(y-1)^2=((x-y-3)/sqrt2)^2=(x-y-3)^2/2`

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

or `2x^2-4x+2+2y^2-4y+2=x^2+y^2+9-6x+6y-2xy`

or `x^2+2xy+y^2+2x-10y-5=0`

graph{(x^2+2xy+y^2+2x-10y-5)((x-1)^2+(y-1)^2-0.03)(x-y-3)=0 [-9.67, 10.33, -3.4, 6.6]}