Let $ P = (2\cos\theta, 2\sin\theta) $ $ \therefore $ coordinates of $ Q = (4\cos\theta + 1,; 6\sin\theta + 3) $ $ \therefore $ locus of $ Q $ is $ \left( \frac{x - 1}{4} \right)^2 + \left( \frac{y - 3}{6} \right)^2 = 1 $ $ e^2 = 1 - \frac{16}{36} = \frac{5}{9} $ $ \Rightarrow \frac{5}{e^2} = 9 $