If $x^3+y^3=3 a x y$, then at $\left(\frac{3 a}{2}, \frac{3 a}{2}\right)$ the value of $3 a y^{\prime \prime}+40$ is

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Aicharya · Accepted Answer

$x^3+y^3=3 a x y$ $3 x^2+3 y^2 \frac{d y}{d x}=3 a y+3 a x \frac{d y}{d x}$ $\begin{aligned} & \Rightarrow\left(y^2-a xight) \frac{d y}{d x}=a y-x^2 \ & \Rightarrow \frac{d y}{d x}=\frac{a y-x^2}{y^2-a x} \Rightarrow\left(\frac{d y}{d x}ight)\left(\frac{3 a}{2}, \frac{3 a}{2}ight)=-1 \ & \Rightarrow\left(2 y \frac{d y}{d x}-aight) \frac{d y}{d x}+\left(y^2-a xight) \frac{d^2 y}{d x^2}=a \frac{d y}{d x}-2 x \ & \Rightarrow 2 y\left(\frac{d y}{d x}ight)^2+\left(y^2-a xight) \frac{d^2 y}{d x^2}=2 a \frac{d y}{d x}-2 x\end{aligned}$ At $\left(\frac{3 a}{2}, \frac{3 a}{2}ight) ; 2 	imes \frac{3 a}{2}(-1)^2+\left(\frac{3 a^2}{4}ight)\left(\frac{d^2 y}{d x^2}ight)$ $=2 a(-1)-3 a$ $\begin{aligned} & \Rightarrow 3 a+\frac{3 a^2}{4}\left(\frac{d^2 y}{d x^2}ight)=-5 a \ & \Rightarrow \frac{3 a}{4}\left(\frac{d^2 y}{d x^2}ight)=-8 \ & \Rightarrow 3 a y^{\prime \prime}=-32 \ & \Rightarrow 3 a y^{\prime \prime}+40=40-32=8\end{aligned}$

If $x^3+y^3=3 a x y$, then at $\left(\frac{3 a}{2}, \frac{3 a}{2}\right)$ the value of $3 a y^{\prime…

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