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Secant of a Curve y =f(x)

For the curve...

Question

For the curve $x = t^{2} - 1, y = t^{2} - t$ , tangent is parallel to $x$ - axis where,

t = 0

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t = \frac{1}{\sqrt{3}}

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t = \frac{1}{2}

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t = - \frac{1}{\sqrt{3}}

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Solution

The correct option is B $t = \frac{1}{2}$
$x = t^{2} - 1, y = t^{2} - t$
If tangent is parallel to $x$ axis, $\frac{d y}{d x} = 0$
$\Rightarrow \frac{d y}{d t} \times \frac{d t}{d x} = 0$
i.e. $(2 t - 1) \times \frac{1}{2 t} = 0$
i.e. $t = \frac{1}{2}$

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