Question

If \(x^2+y^2=t-\frac{1}{t}\) and \(x^4+y^4=t^2+\frac{1}{t^2}\), then \(\frac{x}{y}\frac{dy}{dx}\) is equal to

• A. \(1\)
• B. \(t\)
• C. \(-1\)
• D. \(-t\)

Hint:

When equations contain \(x^2+y^2\) and \(x^4+y^4\), use \[ (x^2+y^2)^2=x^4+y^4+2x^2y^2. \]

Answer 1

The Correct Option is C

Concept:
This question is based on eliminating the parameter and then differentiating implicitly. Given: \[ x^2+y^2=t-\frac{1}{t} \] and \[ x^4+y^4=t^2+\frac{1}{t^2} \] We know: \[ (x^2+y^2)^2=x^4+y^4+2x^2y^2 \]

Step 1: Square the first equation.
\[ (x^2+y^2)^2=\left(t-\frac{1}{t}\right)^2 \] \[ x^4+y^4+2x^2y^2=t^2-2+\frac{1}{t^2} \]

Step 2: Use the second equation.
Given: \[ x^4+y^4=t^2+\frac{1}{t^2} \] Substitute this into the squared equation: \[ t^2+\frac{1}{t^2}+2x^2y^2=t^2-2+\frac{1}{t^2} \] Cancel common terms: \[ 2x^2y^2=-2 \] \[ x^2y^2=-1 \] \[ (xy)^2=-1 \]

Step 3: Differentiate the relation.
Since: \[ x^2y^2=-1 \] Differentiate both sides with respect to \(x\): \[ \frac{d}{dx}(x^2y^2)=0 \] Using product rule: \[ 2xy^2+2x^2y\frac{dy}{dx}=0 \] Divide by \(2xy\): \[ y+x\frac{dy}{dx}=0 \] So: \[ x\frac{dy}{dx}=-y \]

Step 4: Find the required expression.
Divide both sides by \(y\): \[ \frac{x}{y}\frac{dy}{dx}=-1 \] Hence, the correct answer is: \[ \boxed{(C)\ -1} \]