Coordinate Geometry – Ellipse – JEE Main 24 Jan 2026 Shift 2

Solution:

Let point $P(h, k)$ be on $x^2 + y^2 = 4$.
We can parameterize $P$ as $(2\cos\theta, 2\sin\theta)$.
So, $h = 2\cos\theta$ and $k = 2\sin\theta$.

Let point $Q(x, y)$ be the point on the ellipse.
$x = 2h + 1 = 4\cos\theta + 1 \Rightarrow \cos\theta = \frac{x-1}{4}$
$y = 3k + 2 = 6\sin\theta + 2 \Rightarrow \sin\theta = \frac{y-2}{6}$

Using $\cos^2\theta + \sin^2\theta = 1$:
$$\frac{(x-1)^2}{16} + \frac{(y-2)^2}{36} = 1$$
This is an equation of an ellipse with $a^2 = 16$ and $b^2 = 36$.
Since $b > a$, it is a vertical ellipse.

Eccentricity formula: $e^2 = 1 – \frac{a^2}{b^2}$
$$e^2 = 1 – \frac{16}{36} = 1 – \frac{4}{9} = \frac{5}{9}$$

Value required:
$$\frac{5}{e^2} = \frac{5}{5/9} = 9$$

Ans. (9)