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 $A\equiv \left(a t_{1}^{2} , 2 a t_{1}\right)$ , $B\equiv \left(a t_{2}^{2} , 2 a t_{2}\right)$
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, $O A \perp O B$, therefore
Now, slope of line $AB$ is
$\frac{2 a t_{2} - 2 a t_{1}}{a t_{2}^{2} - a t_{1}^{2}}=\frac{2 a \left(t_{2} - t_{1}\right)}{a \left(t_{2} - t_{1}\right) \left(t_{2} + t_{1}\right)}=\frac{2}{\left(t_{2} + t_{1}\right)}$
$\therefore m=\frac{2}{t_{1} + t_{2}}=-t_{1}$
On solving, we get
$t_{1}=\pm\sqrt{2}$
Hence, slope is $+-\sqrt{2}$ .