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

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

1. A

0

2. B

1

3. C

$-1$

4. D

2

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### Solution:

${x}^{2}+{y}^{2}=t-\frac{1}{t},\text{\hspace{0.17em}\hspace{0.17em}}{x}^{4}+{y}^{4}={t}^{2}+\frac{1}{{t}^{2}}$

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

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

${x}^{2}{y}^{2}=-1\cdots \cdots \left(1\right)$

Differentiate with respect to $\text{'}x\text{'}$

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

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

$⇒{x}^{3}y.{y}^{|}=-{\left(xy\right)}^{2}$  [from (1)]

$=-\left(-1\right)$

$=1$

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