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

Nov 30, 2017

Equation of parabola is ${x}^{2} + 2 x y + {y}^{2} + 2 x - 10 y - 5 = 0$

#### 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 $\left(x , y\right)$. Its distance from focus $\left(1 , 1\right)$ is

$\sqrt{{\left(x - 1\right)}^{2} + {\left(y - 1\right)}^{2}}$

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

$| \frac{x - y - 3}{\sqrt{{1}^{2} + {1}^{2}}} | = | \frac{x - y - 3}{\sqrt{2}} |$

Hence equation of parabola is

$\sqrt{{\left(x - 1\right)}^{2} + {\left(y - 1\right)}^{2}} = | \frac{x - y - 3}{\sqrt{2}} |$ and squaring

${\left(x - 1\right)}^{2} + {\left(y - 1\right)}^{2} = {\left(\frac{x - y - 3}{\sqrt{2}}\right)}^{2} = {\left(x - y - 3\right)}^{2} / 2$

or $2 \left({\left(x - 1\right)}^{2} + {\left(y - 1\right)}^{2}\right) = {\left(x - y - 3\right)}^{2}$

or $2 {x}^{2} - 4 x + 2 + 2 {y}^{2} - 4 y + 2 = {x}^{2} + {y}^{2} + 9 - 6 x + 6 y - 2 x y$

or ${x}^{2} + 2 x y + {y}^{2} + 2 x - 10 y - 5 = 0$

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]}