Question

A normal chord $AB$ of a parabola $y^2-12x=0$ subtends a right angle at the vertex of the parabola. If the point of intersection of the normals drawn at $A$ and $B$ is $(p,q)$, then the value of $\frac{p^2}{q^2}$ is

• A. $2$
• B. $1$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is A $2$


$y^2=12x;y^2=4ax\Rightarrow a=3\Rightarrow A:(3t_1^2,6t_1),B:(3t_2^2,6t_2)$

$OA$ is perpendicular to $OB$
$\Rightarrow m_{OA}\times m_{OB}=-1\Rightarrow \frac{6t_1}{3t_1^2}\times \frac{6t_2}{3t_2^2}=-1\Rightarrow t_1{t}_2=-4\cdots \left(1\right)$

If $AB$ is normal at $A$ and intersects the parabola at $B$, then
$t_2=-t_1-\frac{2}{t_1}\Rightarrow t_1{t}_2=-t_1^2-2\Rightarrow t_1=\pm \sqrt{2}\left[\text{From (1)}\right]\Rightarrow t_2=\mp 2\sqrt{2}$

$B$ is the point of intersection of the two normals at $A$ and $B$
Coordinates of $B$ is $(24,\mp 12\sqrt{2})=(p,q)$
$\Rightarrow \frac{p^2}{q^2}=2$

If $AB$ is normal at $B$ and intersects the parabola at $A$, then
$t_1=-t_2-\frac{2}{t_2}\Rightarrow t_1{t}_2=-t_1^2-2\Rightarrow t_2=\pm \sqrt{2}\Rightarrow t_1=\mp 2\sqrt{2}$

$A$ is the point of intersection of the normals at $A$ and $B$
Coordinates of $A$ is $(24,\mp 12\sqrt{2})$
$\Rightarrow \frac{p^2}{q^2}=2$