Question

The length of the normal chord which subtends an angle of $9 0^{o}$ at the vertex of the parabola $y^{2}=4x$ is

• A. $6\sqrt{3}$ units
• B. $7\sqrt{2}$ units
• C. $8\sqrt{2}$ units
• D. $9\sqrt{2}$ units

Answer 1

Answer: (A)

For the normal chord, $t_{2}=-t_{1}-\frac{2}{t_{1}}$ … $\left(1\right)$
Also, the chord subtends an angle of $9 0^{o}$ at the vertex
$\therefore t_{1}t_{2}=-4$ … $\left(2\right)$
From $\left(1\right),t_{1}t_{2}=-t_{1}^{2}-2$ or $-4=t_{1}^{2}-2$
$\therefore t_{1}^{2}=2$
$\therefore t_{2}=-\frac{4}{t_{1}}=-\frac{4}{\sqrt{2}}=-2\sqrt{2}$ or $t_{2}^{2}=8$
Now, $PQ^{2}=a^{2}\left(t_{1}^{2} - t_{2}^{2}\right)^{2}+4a^{2}\left(t_{1} - t_{2}\right)^{2}$
$=1\cdot \left[\left(2 - 8\right)^{2} + 4 \left(\sqrt{2} + 2 \sqrt{2}\right)^{2}\right]=\left[36 + 72\right]$
$PQ^{2}=108$ or $PQ=6\sqrt{3}$ units