Question

If a normal chord at a point $t$ on the parabola $y^{2}=4 \,a \,x$ subtends a right angle at the vertex, then $t$ equals to

• A. 1
• B. $\sqrt{2}$
• C. 2
• D. $\sqrt{3}$

Answer 1

Answer: (B)

$\sqrt{2}$

Answer 2

The perpendicular of the normal to the parabola $y^{2}=4 a x$ at $P$ is Suppose, it meets the parabola at $Q$. If $O$ be the vertex of the parabola, then the combined equation of $O P$ and $O Q$ is a homogeneous equation of second degree. $y^{2}=4 a x\left(\frac{y+t x}{2 a t+a t^{3}}\right)$ $\Rightarrow y^{2}\left(2 a t+a t^{3}\right)=4 a x(y+t x)$ $\Rightarrow 4 a t x^{2}+4 a x y-\left(2 a t+a t^{3}\right) y^{2}=0$ Since, $O P$ and $O Q$ are at right angles, then Coefficient of $x^{2}+$ Coefficient of $y^{2}=0$ $\therefore 4 a t-2 a t-a t^{3}=0$ $\Rightarrow t^{2}=2$ $\Rightarrow t=\sqrt{2}$