If frac{x^2}{a^2} + frac{y^2}{b^2} = 1,then f | vedclass

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(D) Given the equation of the ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Differentiating both sides with respect to $x$: $\frac{2x}{a^2} + \fra

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

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(D) Given the equation of the ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Differentiating both sides with respect to $x$: $\frac{2x}{a^2} + \frac{2y}{b^2} \cdot \frac{dy}{dx} = 0$. $\frac{y}{b^2} \cdot \frac{dy}{dx} = -\frac{x}{a^2} \implies \frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$. Now,differentiate again with respect to $x$ using the quotient rule: $\frac{d^2 y}{dx^2} = -\frac{b^2}{a^2} \cdot \frac{d}{dx} \left( \frac{x}{y} \right) = -\frac{b^2}{a^2} \cdot \left( \frac{y(1) - x(dy/dx)}{y^2} \right)$. Substitute $\frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$: $\frac{d^2 y}{dx^2} = -\frac{b^2}{a^2} \cdot \left( \frac{y - x(-b^2 x / a^2 y)}{y^2} \right) = -\frac{b^2}{a^2} \cdot \left( \frac{a^2 y^2 + b^2 x^2}{a^2 y^3} \right)$. Since $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $b^2 x^2 + a^2 y^2 = a^2 b^2$. Substituting this into the expression: $\frac{d^2 y}{dx^2} = -\frac{b^2}{a^2} \cdot \left( \frac{a^2 b^2}{a^2 y^3} \right) = -\frac{b^4}{a^2 y^3}$.

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

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