If x2 - y2 - t - 1-t and x4 - y4 - t2 - 1-t2- then x3y dy-dx equals

X^2 + y^2 = t 1/t On squaring both the sides = x^4 + y^4 + 2x^2y^2 = t^2 + 1/t^2 2 Given that x^4 + y^4 = t^2 + 1/t^2 ∴ x^2y^2 = 1 ⇒ x^2 .2y dy/dx + y ...

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Byju's Answer

Standard XIII

Mathematics

Implicit Differentiation

If x2 + y2 = ...

Question

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

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Solution

The correct option is C 1
$x^{2} + y^{2} = t - \frac{1}{t}$
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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Similar questions

Q. 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}

equals

Q.
lf

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}

equals

Q. 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} =

x^{2} + y^{2} = t + \frac{1}{t}, x^{4} + y^{4} = t^{2} + \frac{1}{t^{2}} \Rightarrow x^{3} y \frac{d y}{d x} =

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

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

The correct option is C 1x2+y2=t−1t On squaring both the sides =x4+y4+2x2y2 = t2+1t2−2 Given that - x4+y4=t2+1t2 ∴x2y2=−1 ⇒x2.2ydydx+y2.2x=0 ⇒x3ydydx=−x2y2=1

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