If $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then $\frac{d^2y}{dx^2}$ is equal to

$-\frac{b^4}{a^2y^3}$

The correct answer is Option (2) → $-\frac{b^4}{a^2y^3}$

Given ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

Differentiating implicitly w.r.t $x$:

$\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$

Differentiating again to find $\frac{d^2y}{dx^2}$:

$\frac{d^2y}{dx^2} = \frac{d}{dx} \left(-\frac{b^2 x}{a^2 y}\right)$

Use quotient rule or product rule:

$\frac{d^2y}{dx^2} = -\frac{b^2 (y - x \frac{dy}{dx})}{a^2 y^2} = -\frac{b^2 (y - x(-\frac{b^2 x}{a^2 y}))}{a^2 y^2} = -\frac{b^2 (y + \frac{b^2 x^2}{a^2 y})}{a^2 y^2}$

Simplify:

$\frac{d^2y}{dx^2} = -\frac{b^2 (a^2 y^2 + b^2 x^2)}{a^4 y^3} = -\frac{b^2 (b^2 x^2 + a^2 y^2)}{a^4 y^3}$

Using ellipse equation: $b^2 x^2 + a^2 y^2 = a^2 b^2$

Thus, $\frac{d^2y}{dx^2} = -\frac{b^2 \cdot a^2 b^2}{a^4 y^3} = -\frac{b^4}{a^2 y^3}$