If 1-x4+1-y4=a(x2-y2) then dydx=

1+x4+1-y4=ax2-y2 Put x2=sin α y2=sin β 1-sin2α+1+sin2β=a(sin α-sin β) cos α+cos β=a(sin α-sin β) 2cos α+β2 cosα-β2=a2cosα+β2 sin α-β2cos α-β2=a sin α-β2 tan α-β2=1a α-β=2tan-11a sin-1x2-sin-1y2=2tan-1 1a 11-x4(2x)-11-y4(2y)·dydx=0 dydx=x1-y4y1-x4

If 1-x4+1-y4=a(x2-y2) then dydx=

1+x4+1-y4=ax2-y2 Put x2=sin α y2=sin β 1-sin2α+1+sin2β=a(sin α-sin β) cos α+cos β=a(sin α-sin β) 2cos α+β2 cosα-β2=a2cosα+β2 sin α-β2cos α-β2=a sin α-β2 tan α-β2=1a α-β=2tan-11a sin-1x2-sin-1y2=2tan-1 1a 11-x4(2x)-11-y4(2y)·dydx=0 dydx=x1-y4y1-x4

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$If \sqrt{1 - x^{4}} + \sqrt{1 - y^{4}} = a (x^{2} - y^{2}) then \frac{dy}{dx} =$

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$\frac{x}{y} \frac{\sqrt{1 - y^{4}}}{\sqrt{1 - x^{4}}}$

$\frac{x}{y} \frac{\sqrt{1 + y^{2}}}{\sqrt{1 + x^{2}}}$

$\frac{y}{x} \frac{\sqrt{1 - x^{4}}}{\sqrt{1 - y^{4}}}$

$\frac{y}{x} \frac{\sqrt{1 + y^{4}}}{\sqrt{1 + x^{4}}}$

answer is B.

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Detailed Solution

$\sqrt{1 + x^{4}} + \sqrt{1 - y^{4}} = a (x^{2} - y^{2}) Put x^{2} = \sin α y^{2} = \sin β \sqrt{1 - \sin^{2} α} + \sqrt{1 + \sin^{2} β} = a (\sin α - \sin β) \cos α + \cos β = a (\sin α - \sin β) 2 \cos \frac{α + β}{2} \cos \frac{α - β}{2} = a (2 \cos \frac{α + β}{2} \sin \frac{α - β}{2})$

$\cos \frac{α - β}{2} = a \sin \frac{α - β}{2} \tan (\frac{α - β}{2 = \frac{1}{a}}) α - β = 2 \tan^{- 1} \frac{1}{a} \sin^{- 1} x^{2} - \sin^{- 1} y^{2} = 2 \tan^{- 1} \frac{1}{a} \frac{1}{\sqrt{1 - x^{4}}} (2 x) - \frac{1}{\sqrt{1 - y^{4}}} (2 y) \cdot \frac{dy}{dx} = 0 \frac{dy}{dx} = \frac{x \sqrt{1 - y^{4}}}{y \sqrt{1 - x^{4}}}$