Question

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

• A. $\frac{1}{x^2{y}^3}$
• B. $\frac{1}{x{y}^3}$
• C. $\frac{1}{x^2{y}^2}$
• D. $\frac{1}{x^{3}y}$
• E. $\frac{-1}{x^{3}y}$

Answer 1

Answer: (D)

Given, $x^2+y^2=t-\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$
$\Rightarrow$ $x^4+y^4+2x^2{y}^2=t^2+\frac{1}{t^2}-2$
$\Rightarrow$ $x^4+y^4+2x^2{y}^2=x^2+y^4-2$
$\Rightarrow$ $x^2{y}^2+1=0$
$\Rightarrow$ $y^2=\frac{-1}{x^2}$
On differentiating w.r.t. $x,$
we get $2y\frac{dy}{dx}=\frac{2}{x^3}\Rightarrow \frac{dy}{dx}=\frac{1}{x^{3}y}$