Area of the loop formed by the curve given by $x = a (1 - t^{2})$ , $y = a t (1 - t^{2})$ , $- 1 \leq t \leq 1$ , is.

\frac{3 a^{2}}{5}

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\frac{2 a^{2}}{5}

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\frac{5 a^{2}}{8}

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\frac{8 a^{2}}{15}

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Solution

The correct option is D $\frac{8 a^{2}}{15}$
$x = a (1 - t^{2})$
$∴$ $\frac{d x}{d t} = - 2 a t$ , $y = a (1 - t^{3})$
$∴$ $\frac{d y}{d t} = a (1 - 3 t^{2})$
$∴$ $x \frac{d y}{d t} - y \frac{d x}{d t} = a (1 - t^{2}) a (1 - 3 t^{2}) - a (1 - t^{3}) (- 2 a t)$
$∴$ $x \frac{d y}{d t} - y \frac{d x}{d t} = a^{2} (1 - t^{2}) (1 - 3 t^{2}) + 2 a^{2} t^{2} (1 - t^{2})$
$= a^{2} (1 - 2 t^{2} + t^{4})$
$∴$ Required Area $= \frac{1}{2} a^{2} \int_{- 1}^{1} (1 - 2 t^{2} + t^{4}) d t$

$= a^{2} \int_{0}^{1} (1 - 2 t^{2} + t^{4}) d t = a^{2} {(t - \frac{2}{3} t^{3} + \frac{t^{5}}{5})}_{0}^{3}$
$= a^{2} (1 - \frac{2}{3} + \frac{1}{5}) = \frac{8 a^{2}}{15}$

Suggest Corrections

Similar questions

x = a (\frac{1 - t^{2}}{1 + t^{2}})

y = \frac{2 a t}{1 + t^{2}}

x = \frac{2 a t}{1 + t^{2}}

y = a (\frac{1 - t^{2}}{1 + t^{2}})

- 1 \leq t \leq 1

x = t^{2} - 1, y = t^{3} - t

x = t^{2} + t + 1

y = t^{2} - t + 1

y = f (x)

x = \frac{1 - t^{2}}{1 + t^{2}}

y = \frac{2 t}{1 + t^{2}}

t \in R

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