Let the foci and length of the latus rectum of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\ a>b$ be $(\pm5,0)$ and $\sqrt{50}$, respectively. Then, the square of the eccentricity of the hyperbola $\frac{x^2}{b^2}-\frac{y^2}{a^2b^2}=1$ equals ______

JEE Main 2024 · MathsNumerical answerNumerical valueMediumMulti-step

JEE Main 31 January 2024, Shift 1, Maths Q28

Question 28 of 90 in this shift, Maths question 28 of 30, Section B.

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

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\ a>b

(\pm5,0)

and

\sqrt{50}

, respectively. Then, the square of the eccentricity of the hyperbola

\frac{x^2}{b^2}-\frac{y^2}{a^2b^2}=1

equals ______

Official answer

NTA final key.

Chapter: Conic Sections
Topic: Ellipse Eccentricity
Idea tested: Length of Latus Rectum

Same idea in other shifts

Asked 5× in all

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.