Let the ellipse pass through the centre C of the circle of radius r. Let be the focal distances of the point C on the ellipse. Then is equal to [2025]

Question

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 [2025]

Smart Achievers · Accepted Answer

(3)

Given, equation of circle is $x^{2} + y^{2} - 2 x - 4 y - 11 = 0$

It can be written as

${(x - 1)}^{2} + {(y - 2)}^{2} = 16$

$∴$ Centre of circle is (1, 2) and radius is 4.

Now, ellipse $3 x^{2} + p y^{2} = 4$ passes through (1, 2)

$\Rightarrow 3 + 4 p = 4 \Rightarrow p = \frac{1}{4}$

$∴$ Given equation of ellipse is $3 x^{2} + \frac{1}{4} y^{2} = 4$

i.e., $\frac{x^{2}}{4 / 3} + \frac{y^{2}}{16} = 1$

$∴ a^{2} = \frac{4}{3} and b^{2} = 16 \Rightarrow a = \frac{2}{\sqrt{3}} and b = 4$

Now, eccentricity of ellipse $= \sqrt{\frac{1 - a^{2}}{b^{2}}} = \sqrt{1 - \frac{\frac{4}{3}}{16}} = \sqrt{\frac{11}{12}}$

Now, focal distance of ellipse from (1, 2) $= 4 \pm \sqrt{\frac{11}{12}} \times 2$

$∴ f_{1} = 4 + \frac{2 \sqrt{11}}{2 \sqrt{3}} = 4 + \sqrt{\frac{11}{3}} and f_{2} = 4 - \sqrt{\frac{11}{3}}$

Now, $f_{1} f_{2} = 16 - \frac{11}{3} = \frac{37}{3}$

$∴ 6 f_{1} f_{2} - r = 74 - 4 = 70$ .

Solution

Q.
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 [2025]

1 78

2 68

3 70

4 74

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