The angle between the tangents to the curve $y^{2} = 2 a x$ at the points where $x = \frac{a}{2}$ , is

$\frac{π}{6}$

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$\frac{π}{4}$

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$\frac{π}{3}$

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$\frac{π}{2}$

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Solution

The correct option is D
$\frac{π}{2}$

We have, $y^{2} = 2 a x$
Put $x = \frac{a}{2}; y^{2} = 2 a (\frac{a}{2}) \Rightarrow y = \pm a$
$∴$ The points are $(\frac{a}{2}, a)$ and $(\frac{a}{2}, - a)$
On differentiating equarion (1) w.r.t.x, we get
$2 y \frac{d y}{d x} = 2 a \Rightarrow \frac{d y}{d x} = \frac{a}{y}$
At $(\frac{a}{2}, a) \frac{d y}{d x} = \frac{a}{y} = \frac{a}{a} = 1 = m_{1} (s a y) .$
At $(\frac{a}{2}, - a), \frac{d y}{d x} = \frac{a}{y} = \frac{a}{a} = - 1 = m_{2} (s a y) .$
Since $m_{1} m_{2} = - 1$ , the two tangents are at right are at right angles.
Hence (d) is the correct answer.

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