Question

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

• A. $-\dfrac{y}{x}$
• B. $\dfrac{y}{x}$
• C. $\dfrac{x}{2y}$
• D. $-\dfrac{x}{2y}$

Hint:

Try eliminating parameters first before differentiating parametric equations.

Answer 1

The Correct Option is A

Step 1: Squaring the first equation.
\[ (x^2+y^2)^2 = \left(t+\frac{1}{t}\right)^2 \] \[ x^4+y^4+2x^2y^2=t^2+\frac{1}{t^2}+2 \]
Step 2: Using the second equation.
Given \[ x^4+y^4=t^2+\frac{1}{t^2} \] Subtracting, we get \[ 2x^2y^2=2 \Rightarrow x^2y^2=1 \]
Step 3: Differentiating implicitly.
\[ x^2y^2=1 \] \[ 2xy^2+2x^2y\frac{dy}{dx}=0 \]
Step 4: Solving for $\dfrac{dy{dx}$.}
\[ \frac{dy}{dx}=-\frac{y}{x} \]
Step 5: Conclusion.
The value of $\dfrac{dy}{dx}$ is $-\dfrac{y}{x}$.