If x2+y2=2 and y11+A⋅y−3=0 then A =

x2+y2=2differentiating with respect to x2x+2yy1=0y1=−xy⇒y11=−y+xy1y2=−2y3y11+2y3=0−2y3+Ay3=0⇒A=2

x2+y2=2differentiating with respect to x2x+2yy1=0y1=−xy⇒y11=−y+xy1y2=−2y3y11+2y3=0−2y3+Ay3=0⇒A=2

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If $x^{2} + y^{2} = 2$ and $y^{11} + A \cdot y^{- 3} = 0$ then A =

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

differentiating with respect to x

$2 x + 2 y y^{1} = 0$

$y^{1} = - \frac{x}{y}$

$\Rightarrow y^{11} = \frac{- y + x y^{1}}{y^{2}} = - \frac{2}{y^{3}}$

$y^{11} + \frac{2}{y^{3}} = 0$

$- \frac{2}{y^{3}} + \frac{A}{y^{3}} = 0 \Rightarrow A = 2$

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