CUET · MATHS · PYQ PAPER 2025
If \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), then \(\frac{d^2 y}{d x^2}\) is equal to
- A \(\frac{b^4}{a^2 y^3}\)
- B \(-\frac{b^4}{a^2 y^3}\)
- C \(\frac{b^2}{a^4 y^3}\)
- D \(-\frac{b^2}{a^4 y^3}\)
Answer & Solution
Correct Answer
(B) \(-\frac{b^4}{a^2 y^3}\)
Step-by-step Solution
Detailed explanation
\(\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0 \implies \frac{dy}{dx}=-\frac{b^2 x}{a^2 y}\) \(\frac{d^2 y}{d x^2} = -\frac{b^2}{a^2} \frac{d}{dx} \left(\frac{x}{y}\right) = -\frac{b^2}{a^2} \frac{y(1)-x(\frac{dy}{dx})}{y^2}\)…
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