If the normal chord drawn at $(2 a, 2 a \sqrt{2})$ on the parabola $y^2=4 a x$ subtends an angle $\theta$ at…
If the normal chord drawn at $(2 a, 2 a \sqrt{2})$ on the parabola $y^2=4 a x$ subtends an angle $\theta$ at its vertex, then $\theta=$
$45^{\circ}$
$90^{\circ}$
$135^{\circ}$
$60^{\circ}$
Solution
Normal chord is drawn at $P(2 a, 2 a \sqrt{2})$
Let $P Q$ be a normal chord normal at $P$ to the parabola
$\therefore$ Co-ordinates of $Q=(8 a,-4 a \sqrt{2})$
Vertex $=S(0,0)$
Slope of $S P \times$ Slope of $S Q=\frac{2 a \sqrt{2}}{2 a} \times \frac{-4 a \sqrt{2}}{8 a}=-1$
Thus, $\theta=90^{\circ}$.