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

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(A) 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{

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

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(A) 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$$\frac{y}{b^2} \cdot \frac{dy}{dx} = -\frac{x}{a^2}$$\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)$$\frac{d^2 y}{dx^2} = -\frac{b^2}{a^2} \cdot \left( \frac{y(1) - x(\frac{dy}{dx})}{y^2} \right)$Substitute $\frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$:$\frac{d^2 y}{dx^2} = -\frac{b^2}{a^2} \cdot \left( \frac{y - x(-\frac{b^2 x}{a^2 y})}{y^2} \right)$$\frac{d^2 y}{dx^2} = -\frac{b^2}{a^2} \cdot \left( \frac{y + \frac{b^2 x^2}{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 $b^2 x^2 + a^2 y^2 = a^2 b^2$.$\frac{d^2 y}{dx^2} = -\frac{b^2}{a^2} \cdot \left( \frac{a^2 b^2}{a^2 y^3} \right) = -\frac{b^4}{a^2 y^3}$.

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

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