Maximizing area under curve I came across this problem in TMH mathematics for jee.I tried finding the derivative to the curve but I got stuck while evaluating the area of triangle in terms of tangent to the curve.How should I approach ?

"Value of m for which  the area of triangle formed between the axes and any tangent to the  curve $x^my=b^m$ is constant, is? "

 A: The slope $f'_0$ of the tangent line at an arbitrary point $(x_0,b^mx_0^{-m})$ is $f'_0=-mb^mx_0^{-m-1}$.  
Thus, the equation tangent line is given by
$$y=-mb^mx_0^{-m-1}(x-x_0)+b^mx_0^{-m} \tag 1$$
Using $(1)$, we find the $x$-intercept ($x_{int}$) and $y$-intercept ($y_{int}$) are given respectively by
$$\begin{align}
x_{int}&=x_0+\frac{b^mx_0^{-m}}{mb^mx_0^{-m-1}}\\\\
&=\frac{1+m}{m}x_0 \tag 2
\end{align}$$
$$\begin{align}
y_{int}&=b^mx_0^{-m}+mb^mx_0^{-m}\\\\
&=(1+m)b^mx_0^{-m} \tag 3
\end{align}$$
Using $(2)$ and $(3)$, we find that the area $A$ of the triangle of interest is given by
$$\begin{align}
A&=\frac12 \frac{1+m}{m}x_0\,(1+m)b^mx_0^{-m}\\\\
&=\frac12 \frac{(m+1)^2}{m}b^mx_0^{1-m} \tag 4
\end{align}$$
Finally, we can make $A$ is independent of $m$, and thus making $A$ equal to a constant, say $A=C$, by choosing $x_0$ as
$$\bbox[5px,border:2px solid #C0A000]{x_0=\left(2\frac{mb^{-m}}{(m+1)^2}C\right)^{1/(1-m)}}$$
We can also choose a value for $m$ for which $A$ is independent of $x_0$.  Observing the form of $4$, we see that 
$$\bbox[5px,border:2px solid #C0A000]{m=1 \implies A=2b\,\,\text{which is independent of}\,\,x_0}$$
A: The derivation by @Dr.MV is fine but the conclusion is wrong. Why are you solving for $x_{0}$? The question asked for a value of $m$ that makes the area constant (i.e., independent of the tangent point $x_{0}$). I believe the answer is $m=1$, which makes the $x^{1-m}$ term equal to 1.
