JEE Main 2025 Apr 4 Shift 1, Mathematics Q17: Introduction to Ellipse

JEE Main2025Apr 4, Shift 1Mathematics

The length of the latus-rectum of the ellipse, whose foci are $(2, 5)$ and $(2, - 3)$ and eccentricity is $\frac{4}{5}$ , is

Solution

The foci lie on $x = 2$ , so the major axis is vertical; let semi-major axis $= b$ (here $b > a$ ).

Distance between foci: $2 b e = ∣5 - (- 3) ∣ = 8$ , so $b e = 4$ . With $e = \frac{4}{5}$ : $b = 5$ .

For an ellipse with vertical major axis: $a^{2} = b^{2} (1 - e^{2}) = 25 (1 - \frac{16}{25}) = 9$ , so $a = 3$ .

Length of latus rectum $= \frac{2 a ^{2}}{b} = \frac{2 \cdot 9}{5} = \frac{18}{5}$ .

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