\[(x,y)\]
on the curve, the angle between the tangent vector at the point and the \[x\]
axis is three times the angle between the radial vector and the \[x\]
axis. Find the equation of the curve.If
\[\theta\]
is the angle between the radial vector and the \[x\]
axis, then \[\frac{y}{x} = tan \theta\]
then \[tan 3 \theta = \frac{y}{x}\]
.Use
\[tan(A+B)= \frac{tanA+tanB}{1-tanAtanB}\]
/ With \[A=B= \theta\]
, \[tan 2 \theta = \frac{2 tan \theta }{1- tan^2 \theta}\]
and with \[A= \theta , \; B= 2 \theta\]
.\[\begin{equation} \begin{aligned} tan(\theta + 2 \theta ) &= \frac{tan \theta tan2 \theta}{1- tan \theta tan 2 \theta } \\ &= \frac{tan \theta +\frac{2 tan \theta }{1- tan^2 \theta}}{1- tan \theta \frac{2 tan \theta }{1- tan^2 \theta}}= \tan \theta \frac{3-tan^2 \theta}{1-3 tan^2 \theta} \end{aligned} \end{equation}\]
.Then
\[\frac{dy}{dx}= \frac{y}{x} \frac{3-y^2/x^2}{1-3y^2/x^2}= \frac{y}{x} \frac{3x^2-y^2}{x^2-3y^2}\]
.This equation is homogeneous so let
\[y=vx\]
then \[\frac{dy}{dx}= v + x \frac{dv}{dx}\]
and the equation becomes \[v + x \frac{dv}{dx}=v \frac{3-v^2}{1-3v^2} \rightarrow x \frac{dv}{dx}= v \frac{3-v^2}{1-3v^2}-v = \frac{2v(1+v^2)}{1-3v^2}\]
.Separating variables gives
\[\frac{1-3v^2}{v(1+v^2)} dv= \frac{2}{x} dx\]
.We can write this as
\[(\frac{1}{v} - \frac{4v}{1+v^2})dv = \frac{2}{x} dx\]
.Integration gives
\[ln(v)- 2ln(1+v^2)=2 ln(x)+C\]
.Hence
\[ln(\frac{v}{(1+v^2)^2})=ln(Kx^2) \rightarrow \frac{v}{(1+v^2)^2}=Kx^2\]
.Now substitute
\[v= y/x\]
to get \[\frac{y/x}{(1+y^2/x^2)^2}=Kx^2\]
.