If y^2 = ax^2 + bx + c,where a, b, c are cons | vedclass

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(C) Given $y^2 = ax^2 + bx + c$. Differentiating with respect to $x$,we get: $2y \frac{dy}{dx} = 2ax + b$ $\frac{dy}{dx} = \frac{2ax + b}{2y}$ Differe

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a constant

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(C) Given $y^2 = ax^2 + bx + c$. Differentiating with respect to $x$,we get: $2y \frac{dy}{dx} = 2ax + b$ $\frac{dy}{dx} = \frac{2ax + b}{2y}$ Differentiating again with respect to $x$ using the quotient rule: $\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{2ax + b}{2y} \right) = \frac{2a(2y) - (2ax + b)(2 \frac{dy}{dx})}{4y^2}$ Substituting $\frac{dy}{dx} = \frac{2ax + b}{2y}$: $\frac{d^2y}{dx^2} = \frac{4ay - (2ax + b) \frac{2ax + b}{y}}{4y^2} = \frac{4ay^2 - (2ax + b)^2}{4y^3}$ Now,multiply by $y^3$: $y^3 \frac{d^2y}{dx^2} = \frac{4ay^2 - (2ax + b)^2}{4} = \frac{4a(ax^2 + bx + c) - (4a^2x^2 + 4abx + b^2)}{4}$ $y^3 \frac{d^2y}{dx^2} = \frac{4a^2x^2 + 4abx + 4ac - 4a^2x^2 - 4abx - b^2}{4} = \frac{4ac - b^2}{4}$ Since $a, b, c$ are constants,$\frac{4ac - b^2}{4}$ is a constant.

If $y^2 = ax^2 + bx + c$,where $a, b, c$ are constants,then $y^3 \frac{d^2 y}{dx^2}$ is equal to

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