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Two ellipses with the same eccentricity and related latus recta

This type of question, asked 2 times: 29 Jan 2025, 24 Jan 2026 S1. All 2 questions below, oldest first.

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  1. Question 1

    29 Jan 2025

    62
    Let the ellipse, E1:x2a2+y2 b2=1,a>bE_{1}:\frac{x^{2}}{a^{2}}+\frac{y^{2}}{{\text{ }b}^{2}}= 1,a > b and E2:x2 A2+y2 B2=1, A<BE_{2}:\frac{x^{2}}{{\text{ }A}^{2}}+\frac{y^{2}}{{\text{ }B}^{2}}= 1,\text{ }A < B have same eccentricity 13\frac{1}{\sqrt{3}}. Let the product of their lengths of latus rectums be 323\frac{32}{\sqrt{3}}, and the distance between the foci of E1E_{1} be 4. If E1E_{1} and E2E_{2} meet at A,B,CA,B,C and DD, then the area of the quadrilateral ABCD equals:

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  2. Question 2

    24 Jan 2026 S1

    57
    Let each of the two ellipses E1:x2a2+y2 b2=1,(a>b)E_{1}:\frac{x^{2}}{a^{2}}+\frac{y^{2}}{{\text{ }b}^{2}}= 1,(a > b) and E2:x2 A2+y2 B2=1,( A<B)E_{2}:\frac{x^{2}}{{\text{ }A}^{2}}+\frac{y^{2}}{{\text{ }B}^{2}}= 1,(\text{ }A < B) have eccentricity 45\frac{4}{5}. Let the lengths of the latus recta of E1E_{1} and E2E_{2} be l1\mathcal{l}_{1} and l2\mathcal{l}_{2}, respectively, such that 2l12=9l22\mathcal{l}_{1}^{2}= 9\mathcal{l}_{2}. If the distance between the foci of E1E_{1} is 8 , then the distance between the foci of E2E_{2} is

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