Question

If a normal chord at a point $t$ on the parabola $y^2=4ax$ subtends a right angle at the vertex,then $t=$

• A. $1$
• B. $\sqrt{2}$
• C. $2$
• D. $\sqrt{3}$

Answer 1

Answer: (B)

$\sqrt{2}$

A normal to the parabola at $t=t_1$ intersects the parabola again at $t_2=-t_1-\frac{2}{t_1}$

Thus, the two points are $A=(a{t}_1^2,2a{t}_1)$ and $B=(a{t}_2^2,2a{t}_2)$
slopes of lines $AO$ and $BO$ are $\frac{2}{t_1}$and $\frac{2}{t_2}$

Since they make a right angle at the vertex, i.e. $(0,0)$, we get

$\frac{2}{t_2}\times \frac{2}{t_1}=-1$

$\Rightarrow t_1{t}_2=-4$

$\Rightarrow t_1(-t_1-\frac{2}{t_1})=-4$

$\Rightarrow t_1^2+2=4$

i.e. $t_1=\sqrt{2}$