Question

If a variable tangent to the curve $x^{2}y=c^3$ makes intercepts $a,b$ on $x$ and $y$ axis respectively, then prove that $a^{2}b=\left(27/4\right)C^3$.

Answer 1

$\begin{array}{c}{x}^{2}y=c^3\\ y=\frac{c^3}{x^2}\\ \frac{dy}{dx}=\frac{-2c^3}{x^3}\end{array}$

The tangent into the point

$\left(x_1,\frac{c^3}{x_1^2}\right)is{y}^1=\frac{-2c^3}{x^3}$

The tangent crossing the point of the $f\left(x\right)$

$y-\frac{c^3}{x_1^2}=\frac{-2c^3}{x_1^2}\left(x-x_1\right)$

a is where tangent cross x-axis

$i.ey=0$

Similarly $b_{1}i.ex=0$

$\begin{array}{c}y=0\Rightarrow a=\frac{3}{2}{x}_1\to \left(1\right)\\ x=0\Rightarrow b=3\frac{c^3}{x_1^2}\to \left(ii\right)\\ From\left(i\right)and\left(ii\right)\\ a^{2}b=\frac{27}{4}{c}^3\end{array}$