Question

The equation of the parabola whose focus is $(-6,-6)$ and vertex is $(-2,2),$ is

• A. $(2x-y{)}^2+104x+148y-124=0$
• B. $(2x-y{)}^2+104x-148y+124=0$
• C. $(2x-y{)}^2-104x+148y+124=0$
• D. $(2x+y{)}^2+104x+148y-124=0$

Answer 1

The correct option is A $(2x-y{)}^2+104x+148y-124=0$


$\because A$ is mid point of $SK,$
hence coordinates of $K\equiv (2,10)$
Now, directrix of the required parabola will be $\perp SK$
Hence equation of the directrix will be
$y-10=\left(\frac{-6+2}{2+6}\right)(x-2)\Rightarrow x+2y=22$

Now for parabola condition $SP=PM$
$\sqrt{(x+6{)}^2+(y+6{)}^2}=\frac{|x+2y-22|}{\sqrt{1^2+2^2}}\Rightarrow 5(x^2+y^2+12x+12y+72)=(x+2y-22{)}^2\Rightarrow 4x^2+y^2-4xy+104x+148y-124=0\Rightarrow (2x-y{)}^2+104x+148y-124=0$