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)

$\sqrt{2}$

Let chord touches the parabola at two pints $P,Q$

$t_2=-t_1-2/t_1$

As it subtends angle of $90$ at origin

Product of slopes at origin is $-1$.

$m1\times m2=-1$

$m1=2a{t}_1/a{t}_1^2$ ,

$m2=2a{t}_2/a{t}_2^2$

$t_1\times t_2=-4$

From first equation, we have

$-4=t_1(-t_1-2/t_1)$

-$4=-t_1^2-2$

$t_1=-\sqrt{2}$

• $t_2=2\sqrt{(}2)$
• $p(a{t}_1^2,2a{t}_1)Q(a{t}_2^2,2a{t}_2)$
• $p(2a,-2\sqrt{(}2)a\left)Q\right(8a,4\sqrt{(}2\left)a\right)$

Slope of chord line is $m=\frac{6\sqrt{2}a}{6a}$

$m=\sqrt{2}$