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If $x^{2} + y^{2} = t - \frac{1}{t}$ and $x^{4} + y^{4} = t^{2} + \frac{1}{t^{2}}$ , then $x^{3} y \frac{d y}{d x} =$

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a

0

b

1

c

- 1

d

2

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detailed solution

Correct option is B

$x^{2} + y^{2} = t - \frac{1}{t}, x^{4} + y^{4} = t^{2} + \frac{1}{t^{2}}$

$x^{4} + y^{4} + 2 x^{2} y^{2} = t^{2} + \frac{1}{t^{2}} - 2$

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

$x^{2} y^{2} = - 1 \dots \dots (1)$

Differentiate with respect to $' x'$

$2 x . y^{2} + x^{2} .2 y . y^{|} = 0$

$x^{2} y y^{|} = - x y^{2}$

$\Rightarrow x^{3} y . y^{|} = - {(x y)}^{2}$ [from (1)]

$= - (- 1)$

$= 1$

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