## 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{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}
.
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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