Mathematics - Application of Derivative Question with Solution | TestHub

$\because x^{a}y=k^a$ $\Rightarrow a\ln x+\ln y=a\ln k$

$\therefore \frac{\mathrm{a}}{\mathrm{x}}+\frac{1}{\mathrm{y}}\frac{\mathrm{dy}}{\mathrm{dx}}=0$

or $\frac{dy}{dx}=-\frac{ay}{x}$

$Y-y=-\frac{ay}{x}(X-x)$

or $\frac{Y}{\left(\frac{ay}{x}\right)}-\frac{x}{a}=-X+x$

or $\frac{\mathrm{X}}{1}+\frac{\mathrm{Y}}{\left(\frac{\mathrm{ay}}{\mathrm{x}}\right)}=\mathrm{x}\left(1+\frac{1}{\mathrm{a}}\right)$

or $\frac{X}{x\left(1+\frac{1}{a}\right)}+\frac{Y}{ay\left(1+\frac{1}{a}\right)}=1$

∴ Area $=\frac{1}{2}\cdot x\left(1+\frac{1}{a}\right)$ ay $\left(1+\frac{1}{a}\right)$

$$\begin{array}{rlr}& =\frac{\mathrm{ax}}{2}\cdot \frac{{\mathrm{k}}^{\mathrm{a}}}{{\mathrm{x}}^{\mathrm{a}}}\cdot {\left(1+\frac{1}{\mathrm{a}}\right)}^2& =\frac{{\mathrm{ak}}^{\mathrm{a}}}{2{\mathrm{x}}^{\mathrm{a}-1}}{\left(1+\frac{1}{\mathrm{a}}\right)}^2\end{array}$$

∵ Area is constant $\therefore a-1=0$ $\therefore \mathrm{a}=1$