Maths Q12 – JEE-MAIN 2025

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

Clue 1 of 3: Identify Circle Properties

First, find the center and radius of the given circle.

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Step 1: Find the center and radius of the circle✦ Active

The equation of the circle is $x^{2} + y^{2} - 2 x - 4 y - 11 = 0$ . Comparing with the general form $(x - h)^{2} + (y - k)^{2} = R^{2}$ , we complete the square:

(x^{2} - 2 x + 1) + (y^{2} - 4 y + 4) - 11 - 1 - 4 = 0

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

Thus, the center of the circle is $C (1, 2)$ and the radius is $r = \sqrt{16} = 4$ .

Step 2: Determine the ellipse equation and its properties○ Expand

The ellipse $3 x^{2} + p y^{2} = 4$ passes through the center $C (1, 2)$ . Substitute these coordinates into the ellipse equation:

3 (1)^{2} + p (2)^{2} = 4

3 + 4 p = 4

4 p = 1 ⟹ p = \frac{1}{4}

The ellipse equation becomes $3 x^{2} + \frac{1}{4} y^{2} = 4$ . To convert to standard form, divide by 4:

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

Since $16 > 4 / 3$ , the major axis is along the y-axis. So, $a^{2} = 4 / 3$ and $b^{2} = 16$ , which means $b = 4$ . The eccentricity $e$ is given by $e = \sqrt{1 - \frac{a^{2}}{b^{2}}}$ :

e = \sqrt{1 - \frac{4 / 3}{16}} = \sqrt{1 - \frac{4}{48}} = \sqrt{1 - \frac{1}{12}} = \sqrt{\frac{11}{12}}

💡 Teacher's Secret Hint

Remember to correctly identify the major and minor axes based on the denominators.

Step 3: Calculate focal distances and the final expression○ Expand

For a point $(x_{0}, y_{0})$ on an ellipse with major axis along the y-axis, the focal distances $f_{1}$ and $f_{2}$ are given by $f_{1} = b - e y_{0}$ and $f_{2} = b + e y_{0}$ . For point $C (1, 2)$ , $y_{0} = 2$ :

f_{1} = 4 - \sqrt{\frac{11}{12}} (2) = 4 - \frac{2 \sqrt{11}}{2 \sqrt{3}} = 4 - \frac{\sqrt{11}}{\sqrt{3}} = 4 - \frac{\sqrt{33}}{3}

f_{2} = 4 + \sqrt{\frac{11}{12}} (2) = 4 + \frac{\sqrt{33}}{3}

Now, calculate the product $f_{1} f_{2}$ :

f_{1} f_{2} = (4 - \frac{\sqrt{33}}{3}) (4 + \frac{\sqrt{33}}{3}) = 4^{2} - {(\frac{\sqrt{33}}{3})}^{2} = 16 - \frac{33}{9} = 16 - \frac{11}{3} = \frac{48 - 11}{3} = \frac{37}{3}

Finally, calculate $6 f_{1} f_{2} - r$ :

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

💡 Teacher's Secret Hint

Recall the definition of focal distances for an ellipse and how they relate to the major axis and eccentricity.

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Maths Question 12 – JEE-MAIN 2025

Clue 1 of 3: Identify Circle Properties

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