Conic Sections
Conic Section
star_batch_jee_advanced_2025
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 θ between the axes is given by θ = cos^−1√(1/e^2 + 1/e'^2 - 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: The diagonals of the parallelogram are perpendicular when a specific relationship between the eccentricity and semi-major axis is satisfied.
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