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

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
\(2\left(\frac{d y}{d x}\right)^{2}+2 y \frac{d^{2} y}{d x^{2}}=2 a \)
\( \Rightarrow y \frac{d^{2} y}{d x^{2}}=a-\left(\frac{d y}{d x}\right)^{2}\)
\(\Rightarrow y \frac{d^{2} y}{d x^{2}}=a-\left(\frac{2 a x+b}{2 y}\right)^{2}\)
\(\Rightarrow y \frac{d^{2} y}{d x^{2}}=\frac{4 a y^{2}-(2 a x+b)^{2}}{4 y^{2}}\)
\(\Rightarrow 4 y^{3} \frac{d^{2} y}{d x^{2}}=4 a\left(a x^{2}+b x+c\right)\)
\(-\left(4 a^{2} x^{2}+4 a b x+b^{2}\right)\)
\(\Rightarrow 4 y^{3} \frac{d^{2} y}{d x^{2}}=4 a c-b^{2}\)
\(\Rightarrow y^{3} \frac{d^{2} y}{d x^{2}}=\frac{4 a c-b^{2}}{4}=\) constant