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Ellipse through a given point with a given eccentricity: find a and b

The same question, asked 2 times: 27 Jun 2022*, 2 Apr 2026 S1. Every wording below, oldest first.

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  1. Question 1

    27 Jun 2022

    73
    Let the eccentricity of an ellipse x2a2+y2b2=1,a>b\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1,a > b, be 14\frac{1}{4}. If this ellipse passes through the point (−425,3)\left( - 4\sqrt{\frac{2}{5}},3 \right), then a2+b2a^{2}+b^{2} is equal to:

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  2. Question 2

    2 Apr 2026 S1

    60
    Let an ellipse x2a2+y2 b2=1,a<b\frac{x^{2}}{a^{2}}+\frac{y^{2}}{{\text{ }b}^{2}}= 1,a < b, pass through the point (4,3)(4,3) and have eccentricity 53\frac{\sqrt{5}}{3}. Then the length of its latus rectum is :

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