PYQ Vault
EbooksTutor

Ellipse from two given facts, then the eccentricity of a related hyperbola

This type of question, asked 2 times: 31 Jan 2024, 1 Feb 2024. All 2 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    31 Jan 2024

    87
    Let the foci and length of the latus rectum of an ellipse x2a2+y2 b2=1,a>b\frac{x^{2}}{a^{2}}+\frac{y^{2}}{{\text{ }b}^{2}}= 1,a > b be (±5,0)( \pm 5,0) and 50\sqrt{50}, respectively. Then, the square of the eccentricity of the hyperbola x2b2−y2a2b2=1\frac{x^{2}}{b^{2}}-\frac{y^{2}}{a^{2}b^{2}}= 1 equals

  2. Question 2

    1 Feb 2024

    76
    Let x2a2+y2 b2=1,a>b\frac{x^{2}}{a^{2}}+\frac{y^{2}}{{\text{ }b}^{2}}= 1,a > b be an ellipse, whose eccentricity is 12\frac{1}{\sqrt{2}} and the length of the latus rectum is 14\sqrt{14}. Then the square of the eccentricity of x2a2−y2b2=1\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}= 1 is :

    Tap an option to check your answer.