Question

The slope of a chord of the parabola $y^2=4ax$ which is normal at one end and which subtends a right angle at the origin is

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

Answer 1

Answer: (B)

We have,
$y^2=4ax$

Let the chord be $AB$ . Let $A\equiv \left(a{t}_1^2,2a{t}_1\right)$ , $B\equiv \left(a{t}_2^2,2a{t}_2\right)$
Then, slope of $AB$ is $m=-t_1$ .
Slope of line $OA=\frac{2}{t_1}$
Slope of line $OB=\frac{2}{t_2}$
Since, $OA\perp OB$, therefore
Now, slope of line $AB$ is
$\frac{2a{t}_2-2a{t}_1}{a{t}_2^2-a{t}_1^2}=\frac{2a\left(t_2-t_1\right)}{a\left(t_2-t_1\right)\left(t_2+t_1\right)}=\frac{2}{\left(t_2+t_1\right)}$
$\therefore m=\frac{2}{t_1+t_2}=-t_1$
On solving, we get
$t_1=\pm \sqrt{2}$
Hence, slope is $+-\sqrt{2}$ .