Q48 / 75 · JEE Main 2026 (24 January Shift 2) · Mathematics
Let $(h, k)$ lie on the circle $\mathrm{C}: x^{2}+y^{2}=4$ and the point $(2 h+1,3 k+2)$ lie on an ellipse with eccentricity $e$. Then the value of $\frac{5}{e^{2}}$ is equal to $\_\_\_\_$.
Correct answer: 9
Solution
Since $(h, k)$ lies on circle $x^2 + y^2 = 4$, we have $h = 2\cos\theta$ and $k = 2\sin\theta$.
The transformed point is $(2h+1, 3k+2) = (4\cos\theta + 1, 6\sin\theta + 2)$.
Rearranging: $\frac{(x-1)^2}{16} + \frac{(y-2)^2}{36} = 1$
This is an ellipse with $a^2 = 36$, $b^2 = 16$, so $e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{16}{36} = \frac{5}{9}$
Therefore $\frac{5}{e^2} = \frac{5}{5/9} = 9$
The transformed point is $(2h+1, 3k+2) = (4\cos\theta + 1, 6\sin\theta + 2)$.
Rearranging: $\frac{(x-1)^2}{16} + \frac{(y-2)^2}{36} = 1$
This is an ellipse with $a^2 = 36$, $b^2 = 16$, so $e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{16}{36} = \frac{5}{9}$
Therefore $\frac{5}{e^2} = \frac{5}{5/9} = 9$
Hence, the correct answer is 9.