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Ellipse from its eccentricity and a point or focus on it

This type of question, asked 4 times: 2021 Paper 14, 27 Jun 2022, 2 Apr 2026 S1, 5 Apr 2026 S1. All 4 questions below, oldest first.

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  1. Question 1

    2021 Paper 14

    48
    If x2a2+y2b2=1\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1 ellipse passes through (32,1),e=13\left( \sqrt{\frac{3}{2}},1 \right),e=\frac{1}{\sqrt{3}}. Circle centred at one of the focus and radius 23\frac{2}{\sqrt{3}}. These ellipse &\& circle intersect at two points. Find square of the distance between the two points.

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  2. Question 2

    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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  3. Question 3

    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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  4. Question 4

    5 Apr 2026 S1

    59
    Let a focus of the ellipse E:x2a2+y2b2=1E:\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1 be S(4,0)S(4,0) and its eccentricity be 45\frac{4}{5}. If the point P(3,α)P(3,\alpha) lies on E and O is the origin, then the area of △POS\bigtriangleup POS is equal to :

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