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Secant of a Curve y =f(x)

The slope of ...

Question

The slope of normal to the curve $y^{2} = 4 a x$ at a point $(a t^{2}, 2 a t)$ is

\frac{1}{t}

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t

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- t

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- \frac{1}{t}

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Solution

The correct option is D $- t$
$y^{2} = 4 a x$
$\Rightarrow 2 y \frac{d y}{d x} = 4 a \Rightarrow \frac{d y}{d x} = \frac{2 a}{y}$
Therefore, slope of normal to the given curve is, $= {(- \frac{d x}{d y})}_{(a t^{2}, 2 a t)} = - \frac{2 a t}{2 a} = - t$
Hence, option 'C' is correct.

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