Question

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

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

Answer 1

Answer: (B)

$\frac{-y}{x}$

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

Squaring on both sides

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

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

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

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

Differentiate on both sides w.r.t $x$,

$2x{y}^2+2x^{2}y\frac{dy}{dx}=0$

$y+x\frac{dy}{dx}=0$ ...Dividing by $2xy$

$\frac{dy}{dx}=-\frac{y}{x}$