Question

If a chord which is normal to the parabola $y^2=4ax$ at one end subtends a right angle at the vertex, then its slope can be

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

Answer 1

Answer: (A), (B)

The correct options are
A $\sqrt{2}$
B $-\sqrt{2}$

Let $P(a{t}^2,2at)$ be the point where normal is drawn and $Q$ be the other point where normal intersects the curve.
$\therefore$ Q will have $t_1=-t-\frac{2}{t}\dots \left(i\right)$
$\therefore$ Slope of $OP=\frac{2at-0}{a{t}^2-0}=\frac{2}{t}$
Similarly, slope of $OQ=\frac{2}{t_1}$

Right angle subtended at vertex
$\frac{2}{t}\times \frac{2}{t_1}=-1\Rightarrow t{t}_1=-4\dots \left(ii\right)$
From equation $\left(i\right)$ and $\left(ii\right),$ we get
$\frac{-4}{t}=-t-\frac{2}{t}\Rightarrow \frac{2}{t}=t\Rightarrow t=\pm \sqrt{2}$
$\therefore P\equiv (2a,2\sqrt{2}a)$ or $P\equiv (2a,-2\sqrt{2}a)$ and
$Q\equiv (8a,-4\sqrt{2}a)$ or
$Q\equiv (8a,4\sqrt{2}a)$
Now slope of normal when $t=\sqrt{2}$ will be $=\frac{-4\sqrt{2}a-2\sqrt{2}a}{8a-2a}$
$=-\sqrt{2}$
And, slope of normal when $t=-\sqrt{2}$ will be $=\frac{4\sqrt{2}a-(-2\sqrt{2}a)}{8a-2a}$
$=\sqrt{2}$