Question

If $\theta$ be the angle subtended at the focus by the chord which is normal at the point $(\lambda ,\lambda ),\lambda \ne 0$ to the parabola $y^2=4x,$ then the equation of the line making angle $\theta$ with positive $x-$axis and passing through $(1,2)$ is

• A. $y=2$
• B. $x+2y=5$
• C. $x+y=3$
• D. $x=1$

Answer 1

The correct option is D $x=1$

Let $P\equiv (t_1^2,2t_1)$ and $Q\equiv (t_2^2,2t_2)$
Putting $(\lambda ,\lambda )$ in $y^2=4x,$ we get
$\lambda =0,4$
But $\lambda \ne 0$ (Given)
So$,\lambda =4$
$\therefore P\equiv (4,4)\Rightarrow t_1=2$
Also, $t_2=-t_1-\frac{2}{t_1}$
Hence, $t_2=-3$


$\Rightarrow Q\equiv (9,-6)$
Now, Slope of $PF,m_1=\frac{4}{3}$
and Slope of $QF,m_2=-\frac{3}{4}$
$m_1{m}_2=-1$
$\therefore \theta ={90}^{\circ }$
Hence, equation of line passing through $(1,2)$ and inclined at angle $\theta ={90}^{\circ }$ is $x=1$