Maths Q11 – JEE-MAIN 2025

The centre of a circle C is at the centre of the ellipse E:

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

a > b

. Let C pass through the foci

F_{1}

and

F_{2}

of E such that the circle C and the ellipse E intersect at four points. Let P be one of these four points. If the area of the triangle

P F_{1} F_{2}

is 30 and the length of the major axis of E is 17, then the distance between the foci of E is :

Clue 1 of 3: Geometric Interpretation

Relate the given area of the triangle $P F_{1} F_{2}$ to the coordinates of point P and the distance between the foci.

🥷

Ninja Strategy•Geometric Constraint Check

Quickly eliminate options that violate fundamental geometric properties of an ellipse, such as $2 c < 2 a$ .

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Step 1: Set up geometric relationships and given data✦ Active

The center of the ellipse and circle is $(0, 0)$ . The foci of the ellipse are $F_{1} (- c, 0)$ and $F_{2} (c, 0)$ . Since the circle C passes through the foci, its radius is $c$ , and its equation is $x^{2} + y^{2} = c^{2}$ . Let $P (x, y)$ be one of the four intersection points. The base of $△ P F_{1} F_{2}$ is the distance between foci, $F_{1} F_{2} = 2 c$ . The height of the triangle with respect to this base is $| y |$ .

Area of △ P F_{1} F_{2} = \frac{1}{2} \times (2 c) \times | y | = c | y |

Given Area $= 30$ , so $c | y | = 30$ . The length of the major axis is $2 a = 17$ , so $a = \frac{17}{2}$ .

Step 2: Find the y-coordinate of the intersection point P○ Expand

Since P lies on both the circle $x^{2} + y^{2} = c^{2}$ and the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , we can substitute $x^{2} = c^{2} - y^{2}$ from the circle equation into the ellipse equation:

\frac{c^{2} - y^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

Multiply by $a^{2} b^{2}$ to clear denominators:

b^{2} (c^{2} - y^{2}) + a^{2} y^{2} = a^{2} b^{2}

Rearrange terms and use the ellipse property $c^{2} = a^{2} - b^{2}$ (which implies $a^{2} - c^{2} = b^{2}$ ):

b^{2} c^{2} - b^{2} y^{2} + a^{2} y^{2} = a^{2} b^{2} y^{2} (a^{2} - b^{2}) = a^{2} b^{2} - b^{2} c^{2} y^{2} c^{2} = b^{2} (a^{2} - c^{2}) y^{2} c^{2} = b^{2} (b^{2}) y^{2} = \frac{b^{4}}{c^{2}}

Therefore, $| y | = \frac{b^{2}}{c}$ .

💡 Teacher's Secret Hint

Remember the fundamental relationship $c^{2} = a^{2} - b^{2}$ for an ellipse.

Step 3: Calculate

b^{2}

and the distance between foci

2 c

○ Expand

Substitute $| y | = \frac{b^{2}}{c}$ into the area equation $c | y | = 30$ from Step 1:

c (\frac{b^{2}}{c}) = 30 ⟹ b^{2} = 30

Now use the relationship $c^{2} = a^{2} - b^{2}$ with $a = \frac{17}{2}$ and $b^{2} = 30$ :

c^{2} = {(\frac{17}{2})}^{2} - 30 c^{2} = \frac{289}{4} - 30 c^{2} = \frac{289 - 120}{4} c^{2} = \frac{169}{4}

Taking the square root, $c = \frac{13}{2}$ . The distance between the foci is $2 c$ :

2 c = 2 \times \frac{13}{2} = 13

💡 Teacher's Secret Hint

Ensure all values are correctly substituted into the $c^{2} = a^{2} - b^{2}$ formula.

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Maths Question 11 – JEE-MAIN 2025

Clue 1 of 3: Geometric Interpretation

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