Question

Find orthogonal trajectory of x^p +cy^p=1 where p is fixed constant​

Answer 1

Given,

$\small\longrightarrow F:x^p+cy^p=1$

The arbitrary constant c is given by,

$\small\longrightarrow c=\dfrac{1-x^p}{y^p}$

Differentiating our equation wrt x,

$\small\longrightarrow px^{p-1}+cpy^{p-1}\cdot\dfrac{dy}{dx}=0$

Putting the expression of c,

$\small\longrightarrow px^{p-1}+\dfrac{1-x^p}{y^p}\cdot py^{p-1}\cdot\dfrac{dy}{dx}=0$

$\small\longrightarrow x^{p-1}+\dfrac{1-x^p}{y}\cdot\dfrac{dy}{dx}=0$

Now we obtained the first order differential equation of our equation. We need to replace $\small\dfrac{dy}{dx}$ by $\small-\dfrac{dx}{dy}$ in this equation to obtain the orthogonal trajectory.

So,

$\small\longrightarrow x^{p-1}-\dfrac{1-x^p}{y}\cdot\dfrac{dx}{dy}=0$

$\small\text{ \longrightarrow y\ dy=\dfrac{1-x^p}{x^{p-1}}\ dx }$

Integrating both sides,

$\small\displaystyle\longrightarrow\int y\ dy=\int\left(x^{1-p}-x\right)\ dx\quad\dots(1)$

Assume p ≠ 2. Then,

$\small\text{ \longrightarrow\dfrac{y^2}{2}=\dfrac{x^{2-p}}{2-p}-\dfrac{x^2}{2}+c }$

$\small\text{ \longrightarrow\dfrac{x^2+y^2}{2}+\dfrac{1}{(p-2)x^{p-2}}=c }$

If p = 2, then (1) becomes,

$\small\displaystyle\longrightarrow\int y\ dy=\int\left(\dfrac{1}{x}-x\right)\ dx$

$\small\longrightarrow\dfrac{y^2}{2}=\ln|x|-\dfrac{x^2}{2}+c$

$\small\longrightarrow\dfrac{x^2+y^2}{2}-\ln|x|=c$

Hence the orthogonal trajectory is,

$\small\text{ \longrightarrow\underline{\underline{F_{\perp}:\left\{\begin{array}{lr}\dfrac{x^2+y^2}{2}+\dfrac{1}{(p-2)x^{p-2}}=c,&p\neq2\\\\\dfrac{x^2+y^2}{2}-\ln|x|=c,&p=2\end{array}\right.}} }$