Question

If $x^{2}+y^{2}=t+\frac{2}{t}$ and $x^{4}+y^{4}=t^{2}+\frac{4}{t^{2}}$, then $x^{3} y \frac{d y}{d x}$ equals

• A. -1
• B. -2
• C. $\frac{y}{x}$
• D. x y

Answer 1

Answer: (B)

We have,
$x^{2}+y^{2}= t+\frac{2}{t} \text { and } x^{4}+y^{4}=t^{2}+\frac{4}{t^{2}} $
Now, $x^{2}+y^{2}=t+\frac{2}{t}$
On squaring both sides, we get
$x^{4}+y^{4}+2 x^{2} y^{2}=t^{2}+\frac{4}{t^{2}}+4$
$\Rightarrow x^{4}+y^{4}+2 x^{2} y^{2}=x^{4}+y^{4}+4 $
$\Rightarrow 2 x^{2} y^{2}=4$ $\Rightarrow x^{2} y^{2}=2 $
$\Rightarrow y^{2}=\frac{2}{x^{2}}$
Now, differentiating both sides w.r.t. $x$, we get
$2 y \frac{d y}{d x}=\frac{-4}{x^{3}} \Rightarrow x^{3} y \frac{d y}{d x}=-2$