Conic Sections
Conic Section
Allen Star Batch
Grade 11
Question:
If two concentric ellipses are such that the foci of each one are on the other and their major axes are equal. Let $e$ and $e'$ be their eccentricities, then
The quadrilateral formed by joining the foci of the two ellipses is a parallelogram
The angle $\theta$ between the axes is given by $\theta = \cos^{-1}\sqrt{\frac{1}{e^2} + \frac{1}{e'^2} - \frac{1}{e^2e'^2}}$
If $e^2 + e'^2 = 1$, then the angle between the axis of the two ellipses is $90°$
If $e + e' = 1$, then the angle between the axis of the two ellipses is $90°$
Step-by-Step Solution
Key Concept: Two concentric ellipses with equal major axes and foci of each on the other create a symmetric configuration. The relationship between eccentricities e and e' is governed by ae = ae' (equal semi-major axes) and the geometric constraint that foci of one ellipse lie on the other, leading to the condition ae = a'e' and a relationship involving cos²θ = 1 - e² - e'² + e²e'².
Since $O$ is the midpoint of both diagonals $SS'$ and $HH'$, quadrilateral $HSH'S'$ is a parallelogram. With $OH = 2r = OH' = r' = ae'$, point $H$ lies on the auxiliary circle $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (where $b^2 = a^2(1-e^2)$). Using parametric coordinates and the orthogonality condition for the diagonals, we derive $\cos^2\theta = \frac{1}{e^2} + \frac{1}{a^2} - \frac{1}{e^2a^2}$, and for $\theta = 90°$, this yields $e^2 + a^2 = 1$.
Correct Answer: 1,2,3