Coordinate Geometry – Ellipse – JEE Main 28 Jan 2026 Shift 2

Solution:

Given the center $C(1, -2)$.
Since the y-coordinates of the focus and vertex are the same ($-2$), the major axis is horizontal.

Let the semi-major axis be $a$ and the semi-minor axis be $b$.
Distance between Center and Focus = $ae$.
$$ae = \sqrt{(3-1)^2 + (-2+2)^2} = 2$$

Distance between Center and Vertex = $a$.
$$a = \sqrt{(5-1)^2 + (-2+2)^2} = 4$$

We have $a=4$ and $ae=2$.
$$e = \frac{2}{4} = \frac{1}{2}$$

Now, we find $b^2$ using the relation $b^2 = a^2(1-e^2)$.
$$b^2 = 16(1 – \frac{1}{4}) = 16(\frac{3}{4}) = 12$$

Length of Latus Rectum ($LR$) is given by $\frac{2b^2}{a}$.
$$LR = \frac{2(12)}{4} = \frac{24}{4} = 6$$

Ans. (2)