Question

If x² + y² = t – $\frac{1}{t}$ and x⁴ + y⁴ = t² + $\frac{1}{t^{2}}$, then x³y $\frac{dy}{dx}$ equals –

• A. –1
• B. 0
• C. 1
• D. None of these

Answer 1

Answer: (C)

1

Answer 2

x² + y² = t – $\frac{1}{t}$, x⁴ + y⁴ = t² +$\frac{1}{t^{2}}$= $\left( t - \frac{1}{t} \right)^{2}$+ 2

= x⁴ + y⁴ + 2x²y² + 2

\ x²y² = –1

̃ x² . 2y $\frac{dy}{dx}$ + y² . 2x = 0

̃ x³y$\frac{dy}{dx}$ = –x²y² = 1