If (x^2)/(a^2)+(y^2)/(b^2)=1 , then (d^2y)/(dx^2) is equal to

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

By implicit differentiation with respect to x:

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

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

Differentiating again by the quotient rule:

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

Putting in \(\frac{dy}{dx}\):

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

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

From the equation of the ellipse, \(b^2x^2 + a^2y^2 = a^2b^2\):

    \( \displaystyle \frac{d^2y}{dx^2} = -\frac{b^2}{a^2y^3}\cdot\frac{a^2b^2}{a^2} \)

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