JEE Main 2025 April

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: 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 :

Q: Can we use the parametric coordinates of the ellipse to solve this?

Absolutely! You could set $P$ as $(a \cos \theta, b \sin \theta)$. Since $P$ is on the circle $x^2 + y^2 = a^2e^2$, you would have $a^2 \cos^2 \theta + b^2 \sin^2 \theta = a^2e^2$. This is a valid path, but it might involve more trigonometric manipulation. The algebraic substitution method used in the solution is often more direct for this specific problem.

Q: Why does the area formula involve $|y|$?

The area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$. The base is the segment $F_1F_2$ on the $x$-axis, with length $2ae$. The height is the perpendicular distance from $P(x, y)$ to the $x$-axis, which is the absolute value of the $y$-coordinate, $|y|$. Since $P$ could be in any quadrant, we use $|y|$ to ensure the height is positive.

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the geometry. The circle is centered at the origin and passes through the foci of the ellipse. What is the radius of this circle in terms of the ellipse's parameters

a

and

e

? Remember, the distance from the center to a focus is

a e

Hint 2

Hint 2: Use the area of the triangle

P F_{1} F_{2}

. The base of this triangle is the distance between the foci, which is

2 a e

. The height of the triangle is the absolute value of the

y

-coordinate of the intersection point

P (x, y)

. Use the area formula

Area = \frac{1}{2} \times base \times height = 30

to express

∣ y ∣

in terms of

a e

Hint 3

Hint 3: Now, set up the algebraic intersection. Since

P (x, y)

lies on both the circle

x^{2} + y^{2} = (a e)^{2}

and the ellipse

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

, substitute

x^{2} = a^{2} e^{2} - y^{2}

into the ellipse equation. Solve for

y^{2}

and equate it to the expression for

y^{2}

obtained from Hint 2 to find

b^{2}

Visualized Solution

Visualizing the Ellipse $E$

Given Ellipse $E : \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$
Major axis length $2 a = 17$
Therefore, $a = \frac{17}{2}$

The Circle $C$ and Foci

Circle $C$ is centered at $(0, 0)$
$C$ passes through foci $F_{1} (- a e, 0)$ and $F_{2} (a e, 0)$
Radius of circle $R = a e$
Equation of circle $C : x^{2} + y^{2} = (a e)^{2}$

Intersection Point $P$ and Triangle

Let $P (x, y)$ be one of the intersection points.
$P$ lies on circle $C \Rightarrow x^{2} + y^{2} = a^{2} e^{2}$
$P$ lies on ellipse $E \Rightarrow \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$

Area of Triangle $P F_{1} F_{2}$

Base of $△ P F_{1} F_{2} = F_{1} F_{2} = 2 a e$
Height of $△ P F_{1} F_{2} = ∣ y ∣$
Area $= \frac{1}{2} \times (2 a e) \times ∣ y ∣ = 30$
$\Rightarrow a e ∣ y ∣ = 30 \Rightarrow ∣ y ∣ = \frac{30}{a e}$

Setting up the Intersection

From circle: $x^{2} = a^{2} e^{2} - y^{2}$
Substitute into ellipse: $\frac{a ^{2} e ^{2} - y ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$

Simplifying the Expression

$\frac{a ^{2} e ^{2}}{a ^{2}} - \frac{y ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$
$e^{2} + y^{2} (\frac{1}{b ^{2}} - \frac{1}{a ^{2}}) = 1$
$y^{2} (\frac{a ^{2} - b ^{2}}{a ^{2} b ^{2}}) = 1 - e^{2}$

Using Eccentricity Relation

Recall: $b^{2} = a^{2} (1 - e^{2}) \Rightarrow 1 - e^{2} = \frac{b ^{2}}{a ^{2}}$
Also: $a^{2} e^{2} = a^{2} - b^{2}$
Substitute: $y^{2} (\frac{a ^{2} e ^{2}}{a ^{2} b ^{2}}) = \frac{b ^{2}}{a ^{2}}$

Solving for $∣ y ∣$

$y^{2} \frac{e ^{2}}{b ^{2}} = \frac{b ^{2}}{a ^{2}}$
$y^{2} = \frac{b ^{4}}{a ^{2} e ^{2}}$
$∣ y ∣ = \frac{b ^{2}}{a e}$

Equating the two expressions for $∣ y ∣$

From Step 3: $∣ y ∣ = \frac{30}{a e}$
From Step 7: $∣ y ∣ = \frac{b ^{2}}{a e}$
Equating them: $\frac{b ^{2}}{a e} = \frac{30}{a e} \Rightarrow b^{2} = 30$

Finding the distance between foci

Distance between foci $= 2 a e$
$a^{2} e^{2} = a^{2} - b^{2}$
$a^{2} e^{2} = (\frac{17}{2})^{2} - 30$
$a^{2} e^{2} = \frac{289}{4} - 30 = \frac{289 - 120}{4} = \frac{169}{4}$

Final Calculation

$a e = \frac{169}{4} = \frac{13}{2}$
Distance between foci $2 a e = 13$
Final Answer: 13

The Way Forward

Distance between foci $2 a e = 13$
Key Takeaway: The intersection of a circle and ellipse can be solved by substituting $x^{2}$ or $y^{2}$ between equations.
Final Answer: 13

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The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Solution Diagram

Analyzing the Setup

We start with the ellipse

E

defined by the equation:

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

where

a > b

. Given that the major axis length is

17

, we have

2 a = 17

, which implies

a = \frac{17}{2}

The circle

C

is centered at the origin and passes through the foci at

(\pm a e, 0)

. Consequently, the radius of the circle is

R = a e

, and its equation is:

x^{2} + y^{2} = (a e)^{2}

The Triangle of Intersection

Let

P (x, y)

be one of the intersection points of the circle and the ellipse. Connecting

P

to the foci

F_{1}

and

F_{2}

creates a triangle

△ P F_{1} F_{2}

The base of this triangle is the distance between the foci,

F_{1} F_{2} = 2 a e

. The height of the triangle is the vertical distance from

P

to the

x

-axis, given by

∣ y ∣

Given the area of this triangle is

30

, we set up the following equation:

Area = \frac{1}{2} \times (2 a e) \times ∣ y ∣ = 30

This simplifies to:

∣ y ∣ = \frac{30}{a e}

The Algebraic Dance

Since

P

lies on the circle, we have

x^{2} = a^{2} e^{2} - y^{2}

. Substituting this into the ellipse equation yields:

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

Expanding the first term, we obtain

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

. Rearranging the terms gives:

y^{2} (\frac{1}{b ^{2}} - \frac{1}{a ^{2}}) = 1 - e^{2}

Using the identity

1 - e^{2} = \frac{b ^{2}}{a ^{2}}

and

a^{2} - b^{2} = a^{2} e^{2}

, the equation simplifies to:

y^{2} (\frac{a ^{2} - b ^{2}}{a ^{2} b ^{2}}) = \frac{b ^{2}}{a ^{2}} \Rightarrow y^{2} (\frac{a ^{2} e ^{2}}{a ^{2} b ^{2}}) = \frac{b ^{2}}{a ^{2}}

This further reduces to

y^{2} = \frac{b ^{4}}{a ^{2} e ^{2}}

, which means:

∣ y ∣ = \frac{b ^{2}}{a e}

Final Calculation

We now equate the two expressions for

∣ y ∣

\frac{30}{a e} = \frac{b ^{2}}{a e} \Rightarrow b^{2} = 30

To find the distance between the foci (

2 a e

), we use the relation

a^{2} e^{2} = a^{2} - b^{2}

. Substituting

a = \frac{17}{2}

and

b^{2} = 30

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

Taking the square root, we find

a e = \frac{13}{2}

. Therefore, the distance between the foci is:

2 a e = 2 \times \frac{13}{2} = 13

Common Traps & FAQs

Wait, why is the radius of the circle

a e

Great question! The problem states the circle is centered at the origin (the center of the ellipse) and passes through the foci. The coordinates of the foci are

(- a e, 0)

and

(a e, 0)

. The distance from the origin

(0, 0)

to either focus is simply the distance formula:

(a e - 0)^{2} + (0 - 0)^{2} = a e

. Since the circle passes through these points, its radius must be

a e

Can we use the parametric coordinates of the ellipse to solve this?

Absolutely! You could set

P

(a cos θ, b sin θ)

. Since

P

is on the circle

x^{2} + y^{2} = a^{2} e^{2}

, you would have

a^{2} cos^{2} θ + b^{2} sin^{2} θ = a^{2} e^{2}

. This is a valid path, but it might involve more trigonometric manipulation. The algebraic substitution method used in the solution is often more direct for this specific problem.

Why does the area formula involve

∣ y ∣

The area of a triangle is

\frac{1}{2} \times base \times height

. The base is the segment

F_{1} F_{2}

on the

x

-axis, with length

2 a e

. The height is the perpendicular distance from

P (x, y)

to the

x

-axis, which is the absolute value of the

y

-coordinate,

∣ y ∣

. Since

P

could be in any quadrant, we use

∣ y ∣

to ensure the height is positive.

Select Answer:

Visualized Solution

Visualizing the Ellipse $E$

The Circle $C$ and Foci

Intersection Point $P$ and Triangle

Area of Triangle $P F_{1} F_{2}$

Setting up the Intersection

Simplifying the Expression

Using Eccentricity Relation

Solving for $∣ y ∣$

Equating the two expressions for $∣ y ∣$

Finding the distance between foci

Final Calculation

The Way Forward

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Triangle of Intersection

The Algebraic Dance

Final Calculation

Similar Questions

Let $S$ and $S^{'}$ be the foci of the ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$ and $P (α, β)$ be a point on the ellipse in the first quadrant. If $(S P)^{2} + (S^{'} P)^{2} - S P \cdot S^{'} P = 37$ , then $α^{2} + β^{2}$ is equal to :

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

Let $P$ be a variable point on the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ with foci $F_{1}$ and $F_{2}$ . If $A$ is the area of the triangle $P F_{1} F_{2}$ , then the maximum value of $A$ is .........

Select Answer:

Visualized Solution

Visualizing the Ellipse .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E

The Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C and Foci

Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and Triangle

Area of Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PF1​F2​

Setting up the Intersection

Simplifying the Expression

Using Eccentricity Relation

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣y∣

Equating the two expressions for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣y∣

Finding the distance between foci

Final Calculation

The Way Forward

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Triangle of Intersection

The Algebraic Dance

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P be a variable point on the ellipse a2x2​+b2y2​=1 with foci F1​ and F2​. If A is the area of the triangle PF1​F2​, then the maximum value of A is .........

Visualizing the Ellipse $E$

The Circle $C$ and Foci

Intersection Point $P$ and Triangle

Area of Triangle $P F_{1} F_{2}$

Solving for $∣ y ∣$

Equating the two expressions for $∣ y ∣$

Let $P$ be a variable point on the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ with foci $F_{1}$ and $F_{2}$ . If $A$ is the area of the triangle $P F_{1} F_{2}$ , then the maximum value of $A$ is .........