Question
Let (h, k) lie on the circle C : x^ 2 +y^ 2 =4 and the point (2 h+1,3 k+2) lie on an ellipse with eccentricity e. Then the value of 5 e^ 2 is equal to \_\_\_\_.
JEE Main
Mathematics
Ellipse
2026
JEE Main 2026 (24 January Shift 2)
JEE Main Mathematics Question (2026) — Solution
Options
- A. A
- B. B
- C. C
- D. D
Answer
A. A
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
Related: Mathematics — Ellipse · All PYQ Banks