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 \(\mathrm{f}_{1}, \mathrm{f}_{2}\) be the focal distances of the point C on the ellipse. Then \(6 \mathrm{f}_{1} \mathrm{f}_{2}-\mathrm{r}\) is equal to

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Ellipse

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

Answer: (C)

The circle

x^{2} + y^{2} - 2 x - 4 y - 11 = 0

has centre

C (1, 2)

and radius

r = 1 + 4 + 11 = 4

. Since the ellipse

3 x^{2} + p y^{2} = 4

passes through

C

3 + 4 p = 4

, so

p = \frac{1}{4}

and the ellipse is

\frac{x ^{2}}{4/3} + \frac{y ^{2}}{16} = 1

, with major axis along

y

and

b = 4

. The product of the focal distances of a point

(x, y)

b^{2} - e^{2} y^{2}

, where

e^{2} = 1 - \frac{4/3}{16} = \frac{11}{12}

. At

C

y = 2

, so

f_{1} f_{2} = 16 - \frac{11}{12} \cdot 4 = 16 - \frac{11}{3} = \frac{37}{3}

. Therefore

6 f_{1} f_{2} - r = 6 \cdot \frac{37}{3} - 4 = 74 - 4 = 70

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Single correct option