Question

If \(\frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1\), then \(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\) is

• A. \(\frac{-b^{4}}{a}\)
• B. \(\frac{\mathrm{b}^{4}}{\mathrm{a}^{2}}\)
• C. \(\frac{-\mathrm{b}^{4}}{y^{3}}\)
• D. \(\frac{-b^{4}}{a^{2}\cdot y^{3}}\)

Hint:

Differentiate the ellipse equation implicitly twice and simplify using the ellipse equation.

Answer 1

The Correct Option is D

Step 1: Differentiate once
$y' = -\dfrac{b^2x}{a^2y}$.

Step 2: Differentiate again
Apply the quotient rule and replace $y'$ to get a numerator $b^2x^2 + a^2y^2$ over $a^2y^3$, times $-\dfrac{b^2}{a^2}$.

Step 3: Simplify
The numerator equals $a^2b^2$, giving $-\dfrac{b^4}{a^2y^3}$, which matches option D.

Final Answer:
The second derivative is -b^4/(a^2 y^3). \[ \boxed{\text{(D) }\dfrac{-b^{4}}{a^{2}\cdot y^{3}}} \]