Question

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

• A. $0$
• B. $1$
• C. $-1$
• D. none of these

Answer 1

Answer: (B)

$1$

We have $x^2+y^2=t-\frac{1}{t}$ ..........(1)
$x^4+y^4=t^2+\frac{1}{t^2}$ ................(2)
On squaring equation (1), we get
${(x^2+y^2)}^2=t^2+\frac{1}{t^2}-2$
$x^4+2x^2{y}^2+y^4=x^4+y^4-2$ ..............from (2)
$\therefore 2x^2{y}^2=-2$
$x^2{y}^2=-1$
$y^2=-\frac{1}{x^2}$
Differentiating both the sides,
$2y\frac{dy}{dx}=\frac{2}{x^3}$
$x^{3}y\frac{dy}{dx}=1$