Question

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

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is B 1

$x^2+y^2=t-\frac{1}{t}$
$x^4+y^4+2x^2{y}^2=t^2+\frac{1}{t^2}-2$ ...........(1)
$x^4+y^4=x^4+y^4+2x^2{y}^2+2$
$0=x^2{y}^2+1$
$\frac{-1}{x^2}=y^2$

Differentiating both the sides w.r.t x,
$\frac{-1(-2)}{x^3}=2y\frac{dy}{dx}$

$1=x^{3}y\frac{dy}{dx}$