Question

Let tangent and normal to the parabola $y^2=8x$ drawn at $(2,4)$ intersect the line $lx+y=3$ at the points $A$ and $B$ respectively. If $AB$ subtends a right angle at the vertex of the parabola, then the sum of all possible values of $l$ is :

• A. $2$
• B. $1$
• C. $0$
• D. $-1$

Answer 1

Answer: (D)

$-1$

The slopes of the tangent and the normal drawn at $(2,4)$ are $1$ and $-1$ respectively.
Hence, the equation of the normal is $x+y=6$.
The equation of the tangent is $y=x+2$
The point of intersection of the normal $x+y=6$ with $lx+y=3$ is given by $A(\frac{3}{1-l},\frac{3-6l}{1-l})$
The point of intersection of the tangent $y=x+2$ with $lx+y=3$ is given by $B(\frac{1}{1+l},\frac{3+2l}{1+l})$
$AB$ subtends a right angle at the vertex $(0,0)$. Hence, the product of the slopes of $OA$ and $OB$ will be $-1$.
Hence,
$\frac{3+2l}{1}\times \frac{3-6l}{3}=-1$
$3-6l+2l-4l^2=-1$
$\Rightarrow 4l^2+4l-4=0$
$\Rightarrow l^2+l-1=0$
The sum of all values of $l=-1$