Question

The length of the normal chord to the parabola $y^2=4x$ which subtends a right angle at the vertex is

• A. $6\sqrt{3}$
• B. $3\sqrt{3}$
• C. $2$
• D. $1$

Answer 1

The correct option is A $6\sqrt{3}$

Let PQ be the normal chord to the parabola $y^2=4x$ where $a=1$

Let $(a{t}_1^2,2a{t}_1)$ and $(a{t}_2^2,2a{t}_2)$ be the coordinates of P and Q respectively.

i.e., coordinates of $P=(t_1^2,2t_1)$

and coordinates of $Q=(t_2^2,2t_2)$

Equation of normal at point P is

$y-2t_1=\frac{-2t_1}{2}(x-t_1^2)$

$\Rightarrow y-2t_1=-t_1(x-t_1^2)$ …………$\left(1\right)$

Since Q is a point at the normal, substituting $y=2t_2$ and $x=t_2^2$

we have,

$2t_2-2t_1=-t_1(t_2^2-t_1^2)$

$\Rightarrow 2(t_2-t_1)=-t_1(t_2+t_1)(t_2-t_1)$

$\Rightarrow 2=-t_1(t_2+t_1)$……… $\left(2\right)$

It is given that the normal chord subtends a right angle at the vertex $A(0,0)$ i.e., $PA\perp AQ$

$\therefore$ (slope of PA)(slope of AQ)$=-1$

$\left(\frac{(2t_1-0)}{(t_1^2-0)}\right)\left(\frac{2t_2-0}{t_2^2-0}\right)=-1$

$\Rightarrow \frac{2t_1}{t_1^2}\cdot \frac{2t_2}{t_2^2}=-1$

$\Rightarrow \frac{4}{t_1{t}_2}=-1$

$\Rightarrow t_1{t}_2=-4$ ……$\left(2\right)$

From $\left(2\right)$,

$-t_1(t_2+t_1)=2$

$\Rightarrow -t_1{t}_2-t_1^2=2$

$\Rightarrow 4-t_1^2=2$ [from $\left(3\right)$]

$\Rightarrow t_1^2=2$

$\Rightarrow t_1=\sqrt{2}$

Substituting value of $t_1$ in $\left(3\right)$ we have

$t_2=\frac{-4}{\sqrt{2}}$

$t_2=\frac{-4\sqrt{2}}{2}=-2\sqrt{2}$

$\therefore$ coordinates of $P=\left(\right(\sqrt{2}{)}^2,2\left(\sqrt{2}\right))=(2,2\sqrt{2})$

Coordinates of $Q=\left(\right(-2\sqrt{2}{)}^2,2(-2\sqrt{2}))=(-8,-4\sqrt{2})$

$\therefore$ Length of normal chord $PQ=\sqrt{(8-2{)}^2+(-4\sqrt{2}-2\sqrt{2}{)}^2}$

$=\sqrt{6^2+(-6\sqrt{2}{)}^2}$

$=\sqrt{36+72}$

$=\sqrt{108}$

$=6\sqrt{3}$.