Question

The length of normal chord of parabola $y^2=4x$, which subtends an angle of ${90}^{\circ }$ at the vertex is

• A. $6\sqrt{3}\text{units}$
• B. $7\sqrt{2}\text{units}$
• C. $8\sqrt{3}\text{units}$
• D. $9\sqrt{2}\text{units}$

Answer 1

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

$y^2=4x$
$a=1$
Let normal be drawn at point $P(t^2,2t)$and other end of normal be $Q\left(t_0\right)$
Vertex $O=(0,0)$

Slope of $OP=\frac{2t-0}{t^2-0}=\frac{2}{t}$

Slope of $OQ=\frac{2t_0-0}{t_0^2-0}=\frac{2}{t_0}$

$\angle POQ={90}^{\circ }\Rightarrow \frac{2}{t}\cdot \frac{2}{t_0}=-1\Rightarrow t{t}_0=-4\cdots \left(1\right)$
Also,
$t_0=-t-\frac{2}{t}\Rightarrow \frac{-4}{t}=-t-\frac{2}{t}\Rightarrow \frac{2}{t}=t\Rightarrow t^2=2\Rightarrow t=\pm \sqrt{2}\Rightarrow t_0=\mp 2\sqrt{2}$

The coordinates of $P$ are $(2,2\sqrt{2})$ or $(2,-2\sqrt{2})$
And coordinates of $Q$ are $(8,-4\sqrt{2})$ or $(8,4\sqrt{2})$

$\therefore$ Length of $PQ$
$=\sqrt{6^2+(6\sqrt{2}{)}^2}=6\sqrt{3}\text{units}$