Answer : $4x^2+4xy+y^2+4x+30y+16=0$
Given focus of the parabola $S(-1,-2)$ and directrix is $x-2y+3=0$
Let $P(x,y)$ be any point on the parabola
Hence $SP=\sqrt{(x+1)^2+(y+2)^2}$ and length of the perpendicular is $PM=\bigg|\large\frac{x-2y+3}{\sqrt{1^2+(-2)^2}}\bigg|$
But $SP^2=PM^2$
$\therefore (x+1)^2+(y+2)^2=\large\frac{(x-2y+3)^2}{(\sqrt{1^2+(2)^2})^2}$
$\Rightarrow 5[(x+1)^2+(y+2)^2]=x^2+4y^2+9-4xy-12y+6x$
$\Rightarrow 5x^2+10x+5+5y^2+20y+20=x^2+4y^2-4xy-12y+6x$
$\Rightarrow 4x^2+4xy+y^2+4x+30y+16=0$
This is the required equation.