Question

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

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

Answer 1

The correct option is A $\frac{-y}{x}$

We have,

$x+y=t-\frac{1}{t}......\left(1\right)$

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

By equation (2) to,

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

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

$x^2+y^2-2={(t-\frac{1}{t})}^2.......\left(3\right)$

By equation (1) and (3) to, and we get,

$x^2+y^2-2={(x+y)}^2$

$x^2+y^2-2=x^2+y^2+2xy$

$-2=2xy$

$xy=-1$

On differentiating and we get,

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

$x\frac{dy}{dx}+y\left(1\right)=0$

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

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

Hence, this is the answer.