If \(x = a\cos\alpha + b\sin\alpha\) and \(y = a\sin\alpha - b\cos\alpha\), then \(\left(x\displaystyle\frac{dy}{dx} - y^2\frac{d^2y}{dx^2}\right)\) is equal to:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(y\)
Squaring and adding the two equations removes \(\alpha\):
\(x^2 + y^2 = a^2(\cos^2\alpha + \sin^2\alpha) + b^2(\sin^2\alpha + \cos^2\alpha)\)
\(= a^2 + b^2\)
The terms \(\pm 2ab\sin\alpha\cos\alpha\) cancel. By implicit differentiation:
\( \displaystyle 2x + 2y\frac{dy}{dx} = 0,\quad \frac{dy}{dx} = -\frac{x}{y} \)
Differentiating \(x + y\,\frac{dy}{dx} = 0\) once more, by the product rule:
\( \displaystyle 1 + \left(\frac{dy}{dx}\right)^2 + y\frac{d^2y}{dx^2} = 0 \)
\( \displaystyle y\frac{d^2y}{dx^2} = -1 - \frac{x^2}{y^2} = -\frac{x^2 + y^2}{y^2} \)
\( \displaystyle y^2\frac{d^2y}{dx^2} = -\frac{x^2 + y^2}{y} \)
Now
\( \displaystyle x\frac{dy}{dx} - y^2\frac{d^2y}{dx^2} = -\frac{x^2}{y} + \frac{x^2 + y^2}{y} \)
\( \displaystyle = \frac{y^2}{y} = y \)