Question

A tangent and a normal are drawn at the point $P(2,-4)$ on the parabola $y^2=8x$, which meet the directrix of the parabola at the points $A$ and $B$ respectively. If $Q(a,b)$ is a point such that $AQBP$ is a square, then $2a+b$ is equal to

• A. $-18$
• B. $-12$
• C. $-16$
• D. $-20$

Answer 1

The correct option is C $-16$


For the parabola $y^2=8x$,
Directrix is $x=-2$
Slope of tangent at point $P(2,-4)$ is
$\frac{dy}{dx}=\frac{8}{-8}=-1$
$\therefore$ equation of tangent at $P(2,-4)$ is
$y+4=-1(x-2)$
$\Rightarrow x+y+2=0$
Now, for $x=-2\Rightarrow y=0$
$\therefore A\equiv (-2,0)$
Also, normal at $P(2,-4)$ is
$y+4=1(x-2)$
$\Rightarrow x-y=6$
For $x=-2\Rightarrow y=-8$
$\therefore B\equiv (-2,-8)$

Now, since $AQBP$ is a square, therefore mid point of $QP$ is the same as the mid point of $AB$. So, on equating the $x-$ coordinate, we have
$\frac{a+2}{2}=\frac{(-2)+(-2)}{2}$
$\Rightarrow a=-6$
Similarly, for $y-$ coordinate, we have
$\frac{b+(-4)}{2}=\frac{0+(-8)}{2}$
$\Rightarrow b=-4$
$\therefore 2a+b=-16$