1
$\begingroup$

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? "

$\endgroup$
  • $\begingroup$ Find the $ \ x-$ and $ \ y-$ intercepts of the line through the tangent point to the curve: these will give you the base length and height of a right triangle, from which you can calculate the area. You then want to find a value for $ \ m \ $ which eliminates any powers of $ \ x \ $ and $ \ y \ $ from the expression for the area. $\endgroup$ – colormegone Jul 22 '15 at 15:42
2
$\begingroup$

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

$$\bbox[5px,border:2px solid #C0A000]{m=1 \implies A=2b\,\,\text{which is independent of}\,\,x_0}$$

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ That makes it rather trivial then. A constant with respect to what? $\endgroup$ – Mark Viola Jul 22 '15 at 19:41
  • $\begingroup$ How is it trivial? $xy=b$ is a perfectly fine, non-trivial curve. In fact, the answer makes perfect sense geometrically. Since $m=1$ makes the curve symmetric w.r.t. $x$ and $y$ which is why the area is constant. $\endgroup$ – dpmcmlxxvi Jul 22 '15 at 19:43
  • $\begingroup$ A constant with respect to the tangent point. So as you move the tangent up and down the curve the area of the defined triangle is the same. $\endgroup$ – dpmcmlxxvi Jul 22 '15 at 19:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy