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JEE Main202624 January 2026Evening ShiftMathematicsEllipseActual

Let (h, k) lie on the circle C : x²+y²=4 and the point (2 h+1,3 k+2) lie on an ellipse with eccentricity e . Then the value of 5 e² is equal to _ _ _ _ .

Correct answer

0

Step-by-step solution

Since (h, k) lies on circle x^2 + y^2 = 4 , we have h = 2 and k = 2 . The transformed point is (2h+1, 3k+2) = (4 + 1, 6 + 2) . Rearranging: (x-1)^2 16 + (y-2)^2 36 = 1 This is an ellipse with a^2 = 36 , b^2 = 16 , so e^2 = 1 - b^2 a^2 = 1 - 16 36 = 5 9 Therefore 5 e^2 = 5 5/9 = 9

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