If y^ 2 =a x^ 2 +b x+c, where a, b, c are constants, then y^ 3 d^ 2...

Given, $y^{2}=a x^{2}+b x+c$ On differentiating w.r.t. $x$, we get $2 y \frac{d y}{d x}=2 a x+b$ Again differentiating w.r.t. $x$, we get $$\begin{aligned}& 2\left(\frac{d y}{d x}\right)^{2}+2 y \frac{d^{2} y}{d x^{2}}=2 a \\ \Rightarrow \text{ and } y \frac{d^{2} y}{d x^{2}}=a-\left(\frac{d y}{d x}\right)^{2} \end{aligned}$$\Rightarrow \quad y \frac{d^{2} y}{d x^{2}}=a-\left(\frac{2 a x+b}{2 y}\right)^{2}\Rightarrow \quad y \frac{d^{2} y}{d x^{2}}=\frac{4 a y^{2}-(2 a x+b)^{2}}{4 y^{2}}\Rightarrow \quad 4 y^{3} \frac{d^{2} y}{d x^{2}}=4 a\left(a x^{2}+b x+c\right)-\left(4 a^{2} x^{2}+4 a b x+b^{2}\right)\Rightarrow \quad 4 y^{3} \frac{d^{2} y}{d x^{2}}=4 a c-b^{2}\Rightarrow \quad y^{3} \frac{d^{2} y}{d x^{2}}=\frac{4 a c-b^{2}}{4}=$ constant