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Question

Question: Length of the subtangent at \(\left( {{x}_{l}},{{y}_{l}} \right)\) on \({{x}^{n}}{{y}^{m}}={{a}^{m+n...

Length of the subtangent at (xl,yl)\left( {{x}_{l}},{{y}_{l}} \right) on xnym=am+n{{x}^{n}}{{y}^{m}}={{a}^{m+n}} , m,n>0m,n>0 , is.
(a) nmxl\dfrac{n}{m}{{x}_{l}}
(b) mnxl\dfrac{m}{n}\left| {{x}_{l}} \right|
(c) nmyl\dfrac{n}{m}\left| {{y}_{l}} \right|
(d) nmxl\dfrac{n}{m}\left| {{x}_{l}} \right|

Explanation

Solution

Hint:For solving this question first we will simplify the given equation by taking log on both sides then we will differentiate it with respect to xx and calculate the value of dydx\dfrac{dy}{dx} . Then, we will directly find the length of the subtangent from its formula.

Complete step-by-step answer:
Given: We have to find the length of subtangent at (xl,yl)\left( {{x}_{l}},{{y}_{l}} \right) on xnym=am+n{{x}^{n}}{{y}^{m}}={{a}^{m+n}} , m,n>0m,n>0 .
Now, we know that length of subtangent for any curve y=f(x)y=f\left( x \right) at a point (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) on the curve is equal to ydxdy(x1,y1){{\left| y\dfrac{dx}{dy} \right|}_{\left( {{x}_{1}},{{y}_{1}} \right)}} . First, we will solve for dydx\dfrac{dy}{dx} then we will find the length of subtangent using ydxdy(x1,y1){{\left| y\dfrac{dx}{dy} \right|}_{\left( {{x}_{1}},{{y}_{1}} \right)}} .
The equation of the curve is xnym=am+n{{x}^{n}}{{y}^{m}}={{a}^{m+n}} . Now, take log to the base ee on both sides. Then,
xnym=am+n log(xnym)=log(am+n) log(xn)+log(ym)=(m+n)loga nlogx+mlogy=(m+n)loga nlogx+mlogy=(m+n)loga \begin{aligned} & {{x}^{n}}{{y}^{m}}={{a}^{m+n}} \\\ & \Rightarrow \log \left( {{x}^{n}}{{y}^{m}} \right)=\log \left( {{a}^{m+n}} \right) \\\ & \Rightarrow \log \left( {{x}^{n}} \right)+\log \left( {{y}^{m}} \right)=\left( m+n \right)\log a \\\ & \Rightarrow n\log x+m\log y=\left( m+n \right)\log a \\\ & \Rightarrow n\log x+m\log y=( m+n )\log a \\\ \end{aligned}
Now, in the above equation aa is a constant so, differentiating the above equation with respect to xx . Then,
d(nlogx)dx+d(mlogy)dx=d((m+n)loga)dx  \Rightarrow \dfrac{d\left( n\log x \right)}{dx}+\dfrac{d\left( m\log y \right)}{dx}=\dfrac{d\left( \left( m+n \right)\log a \right)}{dx} \\\
We know that (d(logx)dx=1x)(\dfrac{d\left( log x \right)}{dx}={\dfrac{1}{x}})
Then we can write,
nx+mydydx=0 mydydx=nx ymdxdy=xn ydxdy=mnx ydxdy=mnx ydxdy=mnx \Rightarrow \dfrac{n}{x}+\dfrac{m}{y}\dfrac{dy}{dx}=0 \\\ \Rightarrow \dfrac{m}{y}\dfrac{dy}{dx}=-\dfrac{n}{x} \\\ \Rightarrow \dfrac{y}{m}\dfrac{dx}{dy}=-\dfrac{x}{n} \\\ \Rightarrow y\dfrac{dx}{dy}=-\dfrac{m}{n}x \\\ \Rightarrow \left| y\dfrac{dx}{dy} \right|=\left| -\dfrac{m}{n}x \right| \\\ \Rightarrow \left| y\dfrac{dx}{dy} \right|=\dfrac{m}{n}\left| x \right| \\\
Now, from the above calculation, we can say that value of ydxdy(x1,y1){{\left| y\dfrac{dx}{dy} \right|}_{\left( {{x}_{1}},{{y}_{1}} \right)}} for the given curve will be value of mnx\dfrac{m}{n}\left| x \right| at (xl,yl)\left( {{x}_{l}},{{y}_{l}} \right) . Then, length of subtangent =mnxl=\dfrac{m}{n}\left| {{x}_{l}} \right| .

Hence, option (b) is the correct option.

Note: Here if the student directly differentiates the given equation then more calculation will be there. And the student should not confuse the formula of subtangent with the formula of subnormal. Moreover, the student should proceed stepwise and avoid calculation mistakes while solving to get the correct answer.