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