If \(x^2-y^2=t-\displaystyle\frac{1}{t}\), and \(x^4 + y^4 = t^2 +\displaystyle\frac{1}{t^2}\) then which of the following is correct?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(x^3y\displaystyle\frac{dy}{dx}+1=0\)
Instead of differentiating with respect to t, eliminate t. Squaring the first relation:
\( \displaystyle \left(x^2 - y^2\right)^2 = \left(t - \frac{1}{t}\right)^2 \)
\( \displaystyle x^4 + y^4 - 2x^2y^2 = t^2 + \frac{1}{t^2} - 2 \)
Using the second relation, \(x^4 + y^4 = t^2 + \frac{1}{t^2}\), the two sides lose equal terms:
\(-2x^2y^2 = -2\)
\(x^2y^2 = 1\)
Differentiating with respect to x:
\( \displaystyle 2xy^2 + 2x^2y\frac{dy}{dx} = 0 \)
\( \displaystyle \frac{dy}{dx} = -\frac{y}{x} \)
Substituting into Option (4):
\( \displaystyle x^3y\frac{dy}{dx} + 1 = x^3y\left(-\frac{y}{x}\right) + 1 \)
\(= -x^2y^2 + 1 = -1 + 1 = 0\)