Question

Find the isogonal trajectories of the family of curves x²+y²=c if θ=45°.

Answer 1

The given family of curves are

$x^{2}+y^{2}=c$

Differentiating w.r.t. $x$ , we get

$2 x+2 y(d y / d x)=0\\ x+y(d y / d x)=0 ...(i)$

Replacing

$\frac{d y}{d x}\ by \ \frac{\frac{d y}{d x}-\tan 45^{\circ}}{1+\frac{d y}{d x} \tan 45^{\circ}}= \frac{\frac{d y}{d x}-1}{1+\frac{d y}{d x}}$

in Eq. (i), we get

$x+y\left(\frac{\frac{d y}{d x}-1}{1+\frac{d y}{d x}}\right)=0$

 

$\begin{aligned} &x\left(1+\frac{d y}{d x}\right)+y\left(\frac{d y}{d x}-1\right)=0 \\ &\Rightarrow \quad(x+y) \frac{d y}{d x}=(y-x) \\ &\Rightarrow \quad \frac{d y}{d x}=\frac{y-x}{y+x} \quad \ldots(\mathrm{ii}) \end{aligned}$

which is a homogeneous differential equation.

$\begin{aligned} &\text { Put } y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \\ &\Rightarrow \quad v+x \frac{d v}{d x}=\frac{v-1}{v+1} \\ &\Rightarrow \quad x \frac{d v}{d x}=\frac{v-1}{v+1}-v=\frac{v-1-v^{2}-v}{v+1} \\ &\Rightarrow \quad\left(\frac{v+1}{v^{2}+1}\right) d v=-\frac{d x}{x} \end{aligned}$

Integrating, we get

$\begin{aligned} &\frac{1}{2} \log \left|v^{2}+1\right|+\tan ^{-1} v=c_1-\log x \\ &\Rightarrow \quad \log \left|x^{2}+y^{2}\right|+\tan ^{-1}\left(\frac{y}{x}\right)=c_1 \end{aligned}$

Which is the required isogonal trajectories of the given family of curves.