Question

Find the equation of parabola whose
Focus is $(2,3)$ and the directrix is $x-4y+3=0$

Answer 1

Let A(x,y) be the Pt. on the parabola

focus = (2,3) & $e{q}^n$ of directrix

$x-4y+3=0$

Let us draw $\perp$ from A to the directrix

& this $\perp$ be AM (say)

Then, SA = AM

$(SA{)}^2=(AM{)}^2$

$\Rightarrow (x-2{)}^2+(y-3{)}^2={\left|\frac{x-4y+3}{\sqrt{1^2+(-4{)}^2}}\right|}^2$

$\Rightarrow x^2-4x+4+y^2-6y+9=\frac{(x-4y+3{)}^2}{17}$

$\Rightarrow 17x^2+17y^2-68x-102y+221=x^2+16y^2+9-8xy+6x-24y$

$\Rightarrow 16x^2+y^2+8xy-74x-78y+212=0$

Which is the required $e{q}^n$ of

parabola.