JEE Mathematics 2026 Question with Solution

Given: The centre is $(1,-2)$, one focus is $(3,-2)$, and one vertex is $(5,-2)$.

Find: The length of the latus rectum of the ellipse.

Since the centre, focus, and vertex all lie on the line $y=-2$, the major axis is along the $x$-axis.

For an ellipse with major axis along the $x$-axis, the length of latus rectum is

$$\text{Length of latus rectum} = \frac{2b^2}{a}$$

where $a$ is the semi-major axis, $b$ is the semi-minor axis, and $c$ is the distance of the focus from the centre. These satisfy

$$c^2 = a^2 - b^2$$

From the centre $(1,-2)$ and focus $(3,-2)$,

$$c = |3-1| = 2$$

From the centre $(1,-2)$ and vertex $(5,-2)$,

$$a = |5-1| = 4$$

Now use the relation

$$c^2 = a^2 - b^2$$

So,

$$2^2 = 4^2 - b^2$$ $$4 = 16 - b^2$$ $$b^2 = 12$$

Therefore, the length of the latus rectum is

$$L = \frac{2b^2}{a}$$ $$L = \frac{2\times 12}{4}$$ $$L = 6$$

Therefore, the correct option is B.

Given: The ellipse has centre $(1,-2)$, focus $(3,-2)$, and vertex $(5,-2)$.

Find: The latus rectum length.

The equal $y$-coordinate in all three points shows that the major axis is horizontal. Hence the standard parameter meanings are:

• $a$ = distance from centre to vertex
• $c$ = distance from centre to focus
• $b^2 = a^2-c^2$

Now,

$$a = 4, \quad c = 2$$

Thus,

$$b^2 = a^2-c^2 = 16-4 = 12$$

The latus rectum length is

$$\frac{2b^2}{a} = \frac{2\cdot 12}{4} = 6$$

Hence, the latus rectum length is $6$.