If two concentric ellipses be such that the foci of one be on the other and their major axes are equal. Let $e_1$ and $e_2$ be their eccentricities, then prove that the angle between their axes is given by $\cos \theta=\sqrt{\frac{1}{e_1^2}+\frac{1}{e_2^2}-\frac{1}{e_1^2e_2^2}}$.

My approach is as follow $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ represent the ellipse.

Line $y=x\tan \theta$ is the major axis of the second ellipse and its intersection with the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ represent its focus.

With this I am able to find the coordinate of focus but not able to proceed


1 Answer 1


Let the oblique ellipse be

$$\frac{(x\cos \theta-y\sin \theta)^2}{a^2}+\frac{(x\sin \theta+y\cos \theta)^2}{a^2(1-e_2^2)}=1$$

Now $(\pm ae_1,0)$ satisfy the above ellipse, can you proceed with that?


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .