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

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

Differentiating both sides with respect to $x$:

$\frac{d}{dx}\left(\frac{x^2}{a^2}\right) + \frac{d}{dx}\left(\frac{y^2}{b^2}\right) = \frac{d}{dx}(1)$

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

$\frac{2y}{b^2}\cdot\frac{dy}{dx} = -\frac{2x}{a^2}$

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

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Differentiating $\frac{dy}{dx} = -\frac{b^2x}{a^2y}$ with respect to $x$:

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

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

$= -\frac{b^2}{a^2} \cdot \frac{y(1) - x\cdot\frac{dy}{dx}}{y^2}$

$= -\frac{b^2}{a^2} \cdot \frac{y - x\cdot\frac{dy}{dx}}{y^2}$

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Substituting $\frac{dy}{dx} = -\frac{b^2x}{a^2y}$:

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

$= -\frac{b^2}{a^2} \cdot \frac{y + \frac{b^2x^2}{a^2y}}{y^2}$

$= -\frac{b^2}{a^2} \cdot \frac{\frac{a^2y^2 + b^2x^2}{a^2y}}{y^2}$

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

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From $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, multiplying both sides by $a^2b^2$:

$b^2x^2 + a^2y^2 = a^2b^2$

Substituting into the expression:

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

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

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

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