Construction of Curve With Angle of Tangent Vector With x Axis Equal to Three Times Angle of Radial Vector With x Axis

A curve is drawn such that for each point  
\[(x,y)\]
  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  
\[\frac{y/x}{(1+y^2/x^2)^2}=Kx^2\]
.

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