Let the foci and length of the latus rectum of an ellipse { {\x}^2}/{ {\a}^2}+ { {\y}^2}/{ {\b}^2}=1, {\a}> {\b} be ( \pm 5,0) and \sqrt{50}, respectively. Then, the square of the eccentricity of the hyperbola \frac{x^2}{b^2}-\frac{y^2}{a^2 b^2}=1 equals [31-Jan-2024 Shift 1]

Hyperbola

Section: Mathematics

Question : 24 of 92

Marks: +1, -0

Let the foci and length of the latus rectum of an ellipse ‌

+‌

=1,a>b be (±5,0) and √50, respectively. Then, the square of the eccentricity of the hyperbola ‌

−‌

a2b2

=1 equals

[31-Jan-2024 Shift 1]

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