Angle between two quarters of ellipses

Question

I must find the angle between two quarters of ellipses at their common point by the parametric equations:

$R_1(t) = 3\cos (t)i + \sin (t)j$ for $0 \leq t \leq \pi/2$ and

$R2(s) = \cos (s)i + 3\sin (s)j$ for $0 \leq t \leq \pi/2$

To get the point that they intersect, I solved the system of equations:

$3\cos(t)=\cos(s)$ and $\sin(s)=\frac13 \sin(t)$

which after my calculations, I obtained

$\cos(t)=\frac{1}{\sqrt{10}}$, $\sin(t)=\frac{3}{\sqrt{10}}$, $\cos(s)=\frac{3}{\sqrt{10}}$, and $\sin(s)=\frac{1}{\sqrt{10}} $.

Now I know that I have to get the tangents of both $R_1$ and $R_2$, which I can do very easily, but then what do I plug into the new tangent parameters? I am stuck here and would really appreciate any help in solving the rest of the question.

Answer 1

If you have done it your work correctly so far, a far easier method would be to use implicit derivative to find the slope of the tangents at those intersection points. And then you can find the angles separately, then find their difference.

Answer 2

on the ellipse at $R_1,$ the tangent has the slope $$\frac{dy}{dx} = -\frac{\cos t}{3\sin t} = -\frac19 $$ in the same way on $R_2$ we have $$\frac{dy}{dx} = -3\frac{\cos s}{\sin s} = -9 $$ let the angles these tangent make with the positive $x$-axis be $t, s.$ then we have $$\tan t = -1/9, \tan s = -9 \to \tan(t-s) = \frac{\tan t - \tan s}{1+\tan t \tan s}=\frac{9-1/9}{1+1} = \frac49 $$ therefore the angle between the tangent is $$ \tan^{-1}\left(\frac49\right).$$