Question

$x^2+y^2=t+\frac{1}{t},x^4+y^4=t^2+\frac{1}{t^2}\Rightarrow x^{3}y\frac{dy}{dx}=$

• A. $-1$
• B. $1$
• C. $0$
• D. $t$

Answer 1

Answer: (A)

$-1$

Given:

$x^2+y^2=t+\frac{1}{t}-\left(I\right)$ and $x^4+y^4=t^2+\frac{1}{t^2}-\left(II\right)$

By squaring the first equation we get,

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

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

Subtracting Equation $\left(II\right)$ from the above equation, we get

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

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

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

Differentiating this equation with respect to $x$, we get

$2y\frac{dy}{dx}=-\frac{2}{x^3}$

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