If y^2 = a x^2 + 2 x + c,then y^3 frac{d^2 y} | vedclass

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(A) Given the equation $y^2 = a x^2 + 2 x + c$. Differentiating both sides with respect to $x$,we get: $2 y \frac{d y}{d x} = 2 a x + 2$ $y \frac{d y}

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$a c - 1$

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(A) Given the equation $y^2 = a x^2 + 2 x + c$. Differentiating both sides with respect to $x$,we get: $2 y \frac{d y}{d x} = 2 a x + 2$ $y \frac{d y}{d x} = a x + 1$ Differentiating again with respect to $x$ using the product rule: $y \frac{d^2 y}{d x^2} + (\frac{d y}{d x})^2 = a$ From the first derivative,$\frac{d y}{d x} = \frac{a x + 1}{y}$. Substituting this into the second derivative equation: $y \frac{d^2 y}{d x^2} + (\frac{a x + 1}{y})^2 = a$ $y \frac{d^2 y}{d x^2} = a - \frac{(a x + 1)^2}{y^2}$ $y \frac{d^2 y}{d x^2} = \frac{a y^2 - (a x + 1)^2}{y^2}$ Substitute $y^2 = a x^2 + 2 x + c$: $y \frac{d^2 y}{d x^2} = \frac{a(a x^2 + 2 x + c) - (a^2 x^2 + 2 a x + 1)}{y^2}$ $y \frac{d^2 y}{d x^2} = \frac{a^2 x^2 + 2 a x + a c - a^2 x^2 - 2 a x - 1}{y^2}$ $y \frac{d^2 y}{d x^2} = \frac{a c - 1}{y^2}$ Multiplying both sides by $y^2$: $y^3 \frac{d^2 y}{d x^2} = a c - 1$.

If $y^2 = a x^2 + 2 x + c$,then $y^3 \frac{d^2 y}{d x^2}$ is

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