Question

If $x^2+y^2=t+\frac{2}{t}$ and $x^4+y^4=t^2+\frac{4}{t^2}$, then $x^{3}y\frac{dy}{dx}$ 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
$x^4+y^4+2x^2{y}^2=t^2+\frac{4}{t^2}+4$
$\Rightarrow x^4+y^4+2x^2{y}^2=x^4+y^4+4$
$\Rightarrow 2x^2{y}^2=4$ $\Rightarrow x^2{y}^2=2$
$\Rightarrow y^2=\frac{2}{x^2}$
Now, differentiating both sides w.r.t. $x$, we get
$2y\frac{dy}{dx}=\frac{-4}{x^3}\Rightarrow x^{3}y\frac{dy}{dx}=-2$