JEE Main 2025 Apr 8 Shift 2, Mathematics Q10: Ellipse: Chord, Tangent and Normal

JEE Main2025Apr 8, Shift 2Mathematics

Let the ellipse $3 x^{2} + p y^{2} = 4$ pass through the centre $C$ of the circle $x^{2} + y^{2} - 2 x - 4 y - 11 = 0$ of radius $r$ . Let $f_{1}, f_{2}$ be the focal distances of the point $C$ on the ellipse. Then $6 f_{1} f_{2} - r$ is equal to

A
$74$
B
$68$
C
$70$
D
$78$

Solution

Rewrite the circle: $(x - 1)^{2} + (y - 2)^{2} = 16$ , so $C = (1, 2)$ and $r = 4$ .

Rewrite the ellipse in standard form: $\frac{x ^{2}}{4/3} + \frac{y ^{2}}{4/ p} = 1$ .

Since $C = (1, 2)$ lies on the ellipse, $\frac{1}{4/3} + \frac{4}{4/ p} = 1$ , giving $\frac{3}{4} + p = 1$ , so $p = \frac{1}{4}$ .

Then $a^{2} = 4/ p = 16$ (semi-major axis along y) and $b^{2} = 4/3$ (semi-minor along x), confirming a vertical ellipse with $a = 4$ .

Eccentricity: $e^{2} = 1 - \frac{b ^{2}}{a ^{2}} = 1 - \frac{4/3}{16} = 1 - \frac{1}{12} = \frac{11}{12}$ .

For a vertical ellipse, the focal distances of a point $(h, k)$ are $a \pm e k$ ; at $C = (1, 2)$ : $f_{1} = 4 + e \cdot 2$ and $f_{2} = 4 - e \cdot 2$ .

Product: $f_{1} f_{2} = 16 - 4 e^{2} = 16 - 4 \cdot \frac{11}{12} = 16 - \frac{11}{3} = \frac{37}{3}$ .

Therefore $6 f_{1} f_{2} - r = 6 \cdot \frac{37}{3} - 4 = 74 - 4 = 70$ , matching option (3).

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JEE Main 2025 Apr 8 Shift 2, Mathematics Q10: Ellipse: Chord, Tangent and Normal

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