If a variable tangent to the curve ;$x^{2}y=c^{3}$ makes intercepts a, b on x-and y-axes, respectively, then the value of& ;$a^{2}b$ is

Given $x^{2} y=c^{3}$

Let variable point is $\left(x_{1}, y_{1}\right)$ on the curve.

$\begin{array}{l} \left(x^{2}\right)\left(\frac{d y}{d x}\right)+(2 x)(y)=0 \\ x^{2}\left(\frac{d y}{d x}\right)=-2 x y \Rightarrow \frac{d y}{d x}=\frac{-2 x y}{x^{2}}=\frac{-2 y}{x} \\ \text { At }\left(x_{1}, y_{1}\right) \Rightarrow \frac{d y}{d x}=\frac{-2 y_{1}}{x_{1}} \end{array}$

Equation of tangent is,

$\begin{array}{l} y-y_{1}=\frac{-2 y_{1}}{x_{1}}\left(x-x_{1}\right) \\ y x_{1}-2 x y_{1}=3 x_{1} y_{1} \\ \text { using } x_{1}^{2} y_{1}=c^{3} \end{array}$

Equation of tangent becomes,

$y x_{1}^{2}+\frac{2 c^{3} x}{x_{1}}=3 c^{3}$

$\therefore x$-intercept is $a=\frac{3 x_{1}}{2}$

4 -intercept is $b=\frac{3 c^{3}}{x_{1}^{2}}$

$\therefore a^{2} b=\left(\frac{3 x_{1}}{2}\right)^{2} \frac{3 c^{3}}{x_{1}^{2}}=\frac{27 c^{3}}{4}$