Q. If tangents at extremities of a focal chord $AB$ of the parabola $y^{2} = 4ax$ intersect at a point $C$, then $< \text{ACB}$ is equal to

A) $\frac{\pi}{4}$
B) $\frac{\pi}{3}$
C) $\frac{\pi}{2}$
D) $\frac{\pi}{6}$
Explanation

(c)
Let the extremities of focal chords be $A(at_{1}^{2},2at_{1})$ and $B\left( at_{2}^{2},2at_{2} \right)$
The equation of tangents at $A$ and $B$ are
$t_{1y} = x + at_{2}^{2}$ and $t_{2}y = x + at_{2}^{2}$
which meets the points $C$
Slopes of these lines are $m_{1} = \frac{1}{t_{1}\ },m_{2} = \frac{1}{t_{2}}$
Now, $m_{1}m_{2} = \frac{1}{t_{1}} \times \frac{1}{t_{2}}$
$= \frac{1}{- 1}\ \ (\because t_{1}t_{2} = - 1)$
Hence, $\angle ACB = 90{^\circ} = \frac{\pi}{2}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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