Question

Find the equation of parabola whose focus is the point $(-6,-6)$ and vertex is at $(-2,2).$

• A. ${(2x-y)}^2+104x+148y-124=0$
• B. ${(x-2y)}^2-104x-148y-124=0$
• C. ${(2x-y)}^2+52x+74y-62=0$
• D. ${(x-y)}^2+52x+74y-62=0$

Answer 1

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

Given focus $S(-6,-6)$ and vertex $A(-2,2)$
Slope of SA $=\frac{-6-2}{-6+2}=2$
Let $MZ$ be the directrix. Since, directrix is perpendicular to $SA$.
Slope of directrix is $-\frac{1}{2}$
Let the coordinates of the foot of the directrix $Z$ be $(h,k)$.
Since, $A$ is mid-point of $SZ$.
$\therefore \frac{-6+h}{2}=-2,\frac{-6+k}{2}=2$
$\Rightarrow h=2,k=10$
So, the coordinates of $Z$ are $Z(2,10)$
Now, equation of directrix $MZ$ is
$y-10=-\frac{1}{2}(x-2)$
or $x+2y-22=0$
Let $P(x,y)$ be any point on the parabola
By definition $SP=PM$
$\therefore \sqrt{{(x+6)}^2+{(y+6)}^2}=\frac{x+2y-22}{\sqrt{5}}$
Square both sides and simplify, we get
$4x^2+y^2-4xy+104x+148y-124=0$
or ${(2x-y)}^2+104x+148y-124=0$