If \( x^2+y^2=t+\displaystyle\frac{1}{t} \) and \( x^4+y^4=t^2+\displaystyle\frac{1}{t^2} \) , then \( \displaystyle\frac{dy}{dx} \) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( -\displaystyle\frac{y}{x} \)
Squaring the first equation,
\( \displaystyle x^4+y^4+2x^2y^2=t^2+2+\frac{1}{t^2} \)
Subtracting the second equation,
\( \displaystyle 2x^2y^2=2 \)
\( \displaystyle x^2y^2=1 \)
\( \displaystyle xy=\pm 1 \)
So xy is a constant. Differentiating,
\( \displaystyle x\frac{dy}{dx}+y=0 \)
\( \displaystyle \frac{dy}{dx}=-\frac{y}{x} \)