I f x 2 + y 2 = t − 1 t a n d x 4 + y 4 = t 2 + 1 t 2 , t h e n x 3 y d y d x e q u a l s Options: A . &nbsp- 1 B . &nbsp0 C . &nbsp1 D . &nbspNone of these

Answer: Option C : C x 2 + y 2 = t − 1 t On squaring both the sides = x 4 + y 4 + 2 x 2 y 2 = t 2 + 1 t 2 − 2 Given that - x 4 + y 4 = t 2 + 1 t 2 ∴ x 2 y 2 = − 1 ⇒ x 2 .2 y d y d x + y 2 . 2 x = 0 ⇒ x 3 y d y d x = − x 2 y 2 = 1

If x2+y2=t−1t and x4+y4=t2+1t2,then x3ydydx equa

Question

I f x^{2} + y^{2} = t - \frac{1}{t} a n d x^{4} + y^{4} = t^{2} + \frac{1}{t^{2}}, t h e n x^{3} y \frac{d y}{d x} e q u a l s

Options:

A . - 1

B . 0

C . 1

D . None of these

Answer: Option C
:
C

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

On squaring both the sides

= x^{4} + y^{4} + 2 x^{2} y^{2}

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

Given that -

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

∴ x^{2} y^{2} = - 1

\Rightarrow x^{2} .2 y \frac{d y}{d x} + y^{2} . 2 x = 0

\Rightarrow x^{3} y \frac{d y}{d x} = - x^{2} y^{2} = 1

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