If y^{2}=a x^{2}+b x+c,where a, b, c are cons | vedclass

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(A) Given,$y^{2}=a x^{2}+b x+c$. On differentiating with respect to $x$,we get $2 y \frac{d y}{d x}=2 a x+b$,which implies $\frac{d y}{d x} = \frac{2

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a constant

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(A) Given,$y^{2}=a x^{2}+b x+c$. On differentiating with respect to $x$,we get $2 y \frac{d y}{d x}=2 a x+b$,which implies $\frac{d y}{d x} = \frac{2 a x + b}{2 y}$. Again differentiating with respect to $x$,we get $2(\frac{d y}{d x})^{2} + 2 y \frac{d^{2} y}{d x^{2}} = 2 a$. Dividing by $2$,we have $(\frac{d y}{d x})^{2} + y \frac{d^{2} y}{d x^{2}} = a$. Substituting $\frac{d y}{d x} = \frac{2 a x + b}{2 y}$,we get $y \frac{d^{2} y}{d x^{2}} = a - (\frac{2 a x + b}{2 y})^{2}$. $y \frac{d^{2} y}{d x^{2}} = \frac{4 a y^{2} - (2 a x + b)^{2}}{4 y^{2}}$. Substituting $y^{2} = a x^{2} + b x + c$,we get $y \frac{d^{2} y}{d x^{2}} = \frac{4 a (a x^{2} + b x + c) - (4 a^{2} x^{2} + 4 a b x + b^{2})}{4 y^{2}}$. $y \frac{d^{2} y}{d x^{2}} = \frac{4 a^{2} x^{2} + 4 a b x + 4 a c - 4 a^{2} x^{2} - 4 a b x - b^{2}}{4 y^{2}}$. $y \frac{d^{2} y}{d x^{2}} = \frac{4 a c - b^{2}}{4 y^{2}}$. Multiplying both sides by $y^{2}$,we get $y^{3} \frac{d^{2} y}{d x^{2}} = \frac{4 a c - b^{2}}{4}$,which is a constant.

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

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