PYQ Vault
EbooksTutor

Ellipse and hyperbola linked by the product of their eccentricities

This type of question, asked 4 times: 2021 Paper 4, 27 Jul 2022, 27 Jan 2024, 7 Apr 2025. All 4 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    2021 Paper 4

    69
    A hyperbola passes through the foci of the ellipse x225+y216=1\frac{x^{2}}{25}+\frac{y^{2}}{16}= 1 and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is:

    Tap an option to check your answer.

  2. Question 2

    27 Jul 2022

    84
    An ellipse E:x2a2+y2b2=1E:\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1 passes through the vertices of the hyperbola H:x249−y264=−1H:\frac{x^{2}}{49}-\frac{y^{2}}{64}= - 1. Let the major and minor axes of the ellipse E coincide with the transverse and conjugate axes of the hyperbola HH. Let the product of the eccentricities of EE and HH be 12\frac{1}{2}. If ll is the length of the latus rectum of the ellipse EE, then the value of 113l113l is equal to

  3. Question 3

    27 Jan 2024

    168
    Let e1e_{1} be the eccentricity of the hyperbola x216−y29=1\frac{x^{2}}{16}-\frac{y^{2}}{9}= 1 and e2e_{2} be the eccentricity of the ellipse x2a2+y2b2=1,a>b\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1,a > b, which passes through the foci of the hyperbola. If e1e2=1e_{1}e_{2}= 1, then the length of the chord of the ellipse parallel to the xx-axis and passing through (0,2)(0,2) is:

    Tap an option to check your answer.

  4. Question 4

    7 Apr 2025

    143
    Let e1e_{1} and e2e_{2} be the eccentricities of the ellipse x2 b2+y225=1\frac{x^{2}}{{\text{ }b}^{2}}+\frac{y^{2}}{25}= 1 and the hyperbola x216−\frac{x^{2}}{16}- y2 b2=1\frac{y^{2}}{{\text{ }b}^{2}}= 1, respectively. If b<5b < 5 and e1e2=1e_{1}e_{2}= 1, then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is:

    Tap an option to check your answer.