Question

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

• A. $0$
• B. $1$
• C. $-1$
• D. $Noneofthese$

Answer 1

The correct option is B $1$

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

Squaring on both sides

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

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

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

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

$x^2{y}^2=-1$

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

$y=\frac{1}{x}\sqrt{-1}=\frac{i}{x}$

$x^3.y\frac{dy}{dx};\frac{dy}{dx}=\frac{-1}{x^2}i$

$x^3.\frac{i}{x}=\frac{-i}{x^2}$

$=\frac{{-i}^2}{x^3}\times x^3={-i}^2=-(-1)$

$=1$