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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
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The correct answer is:
a constant
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
$$
\begin{aligned}
& 2\left(\frac{d y}{d x}\right)^{2}+2 y \frac{d^{2} y}{d x^{2}}=2 a \\
\Rightarrow & y \frac{d^{2} y}{d x^{2}}=a-\left(\frac{d y}{d x}\right)^{2}
\end{aligned}
$$
$\Rightarrow \quad y \frac{d^{2} y}{d x^{2}}=a-\left(\frac{2 a x+b}{2 y}\right)^{2}$
$\Rightarrow \quad y \frac{d^{2} y}{d x^{2}}=\frac{4 a y^{2}-(2 a x+b)^{2}}{4 y^{2}}$
$\Rightarrow \quad 4 y^{3} \frac{d^{2} y}{d x^{2}}=4 a\left(a x^{2}+b x+c\right)$
$-\left(4 a^{2} x^{2}+4 a b x+b^{2}\right)$
$\Rightarrow \quad 4 y^{3} \frac{d^{2} y}{d x^{2}}=4 a c-b^{2}$
$\Rightarrow \quad y^{3} \frac{d^{2} y}{d x^{2}}=\frac{4 a c-b^{2}}{4}=$ constant
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
$$
\begin{aligned}
& 2\left(\frac{d y}{d x}\right)^{2}+2 y \frac{d^{2} y}{d x^{2}}=2 a \\
\Rightarrow & y \frac{d^{2} y}{d x^{2}}=a-\left(\frac{d y}{d x}\right)^{2}
\end{aligned}
$$
$\Rightarrow \quad y \frac{d^{2} y}{d x^{2}}=a-\left(\frac{2 a x+b}{2 y}\right)^{2}$
$\Rightarrow \quad y \frac{d^{2} y}{d x^{2}}=\frac{4 a y^{2}-(2 a x+b)^{2}}{4 y^{2}}$
$\Rightarrow \quad 4 y^{3} \frac{d^{2} y}{d x^{2}}=4 a\left(a x^{2}+b x+c\right)$
$-\left(4 a^{2} x^{2}+4 a b x+b^{2}\right)$
$\Rightarrow \quad 4 y^{3} \frac{d^{2} y}{d x^{2}}=4 a c-b^{2}$
$\Rightarrow \quad y^{3} \frac{d^{2} y}{d x^{2}}=\frac{4 a c-b^{2}}{4}=$ constant
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