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Friday, 9 October

JEE Main 2022 · MathsNumerical answerNumerical valueHardMulti-step

JEE Main 28 July 2022, Shift 1, Maths Q27

Question 27 of 90 in this shift, Maths question 27 of 30, Section B.

For the hyperbola H:x2−y2=1H:x^2-y^2=1 and the ellipse E:x2a2+y2b2=1, a>b>0E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\ a>b>0, let the (1) eccentricity of E be reciprocal of the eccentricity of H, and (2) the line y=52x+Ky=\sqrt{\frac{5}{2}}x+K be a common tangent of E and H. Then 4(a2+b2)4(a^2+b^2) is equal to __________.

Official answer

3

NTA final key (2022 Session 2).