JEE Main 2026 (23 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: Let $P Q$ be a chord of the hyperbola $\frac{x ^{2}}{4} - \frac{y ^{2}}{b ^{2}} = 1$ , perpendicular to the $x$ -axis such that $O P Q$ is an equilateral triangle, $O$ being the centre of the hyperbola. If the eccentricity of the hyperbola is $3$ , then the area of the triangle $O P Q$ is

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Progressive Hints

Hint 1

Hint 1: Visualize the geometry. Since the chord PQ is perpendicular to the x-axis, the triangle OPQ is symmetric about the x-axis. This means the side PQ is vertical, and the origin O lies on the perpendicular bisector of PQ. What does this imply about the coordinates of P and Q?

Hint 2

Hint 2: Use the equilateral triangle property. If OPQ is equilateral, the distance from the origin O(0,0) to P(x0, y0) must equal the length of the side PQ. Express the length of PQ as 2|y0| and the distance OP as sqrt(x0^2 + y0^2). Square both sides to find a relationship between x0^2 and y0^2.

Hint 3

Hint 3: Now, use the hyperbola equation. You have the relation x0^2 = 3y0^2 from the triangle geometry. Substitute this into the hyperbola equation x0^2/4 - y0^2/8 = 1 to solve for y0^2. Once you have y0^2, the area of an equilateral triangle with side length s is (sqrt(3)/4) * s^2.

Visualized Solution

Hyperbola Equation and Parameters

Given hyperbola: $\frac{x ^{2}}{4} - \frac{y ^{2}}{b ^{2}} = 1$
Standard form: $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1 ⟹ a^{2} = 4$
Eccentricity $e = 3$

Eccentricity Formula

Eccentricity formula: $e^{2} = 1 + \frac{b ^{2}}{a ^{2}}$

Substituting Known Values

Substitute $e = 3$ and $a^{2} = 4$
$3 = 1 + \frac{b ^{2}}{4}$

Solving for $b^{2}$

$\frac{b ^{2}}{4} = 2$
$b^{2} = 8$
Hyperbola equation: $\frac{x ^{2}}{4} - \frac{y ^{2}}{8} = 1$

Defining the Chord $P Q$

Chord $P Q ⊥$ x-axis
Let $P = (x_{0}, y_{0})$ and $Q = (x_{0}, - y_{0})$

Equilateral Triangle Condition

$Δ O P Q$ is equilateral
Therefore, $O P = P Q$

Setting Up the Distance Equation

Distance $O P = x_{0}^{2} + y_{0}^{2}$
Length $P Q = 2 y_{0}$
Condition: $x_{0}^{2} + y_{0}^{2} = 2 y_{0}$

Simplifying the Geometric Condition

Square both sides: $x_{0}^{2} + y_{0}^{2} = 4 y_{0}^{2}$
Rearrange: $x_{0}^{2} = 3 y_{0}^{2}$

Point $P$ on the Hyperbola

Point $P (x_{0}, y_{0})$ lies on the hyperbola
Satisfies equation: $\frac{x _{0}^{2}}{4} - \frac{y _{0}^{2}}{8} = 1$

Substituting $x_{0}^{2}$

Substitute $x_{0}^{2} = 3 y_{0}^{2}$ into the hyperbola equation
$\frac{3 y _{0}^{2}}{4} - \frac{y _{0}^{2}}{8} = 1$

Solving for $y_{0}^{2}$

Multiply by $8$ : $6 y_{0}^{2} - y_{0}^{2} = 8$
$5 y_{0}^{2} = 8 ⟹ y_{0}^{2} = \frac{8}{5}$

Area Formula for $Δ O P Q$

Area of equilateral triangle = $\frac{3}{4} \times (side)^{2}$
Side length $P Q = 2 y_{0}$

Simplifying the Area Expression

Area = $\frac{3}{4} \times (2 y_{0})^{2}$
Area = $3 y_{0}^{2}$

Calculating the Final Area

Substitute $y_{0}^{2} = \frac{8}{5}$
Area = $3 \times \frac{8}{5} = \frac{8 3}{5}$

Final Conclusion

Final Answer: The area of $Δ O P Q$ is $\frac{8 3}{5}$ square units.

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

Solution Diagram

Analyzing the Setup

We are exploring a hyperbola defined by the equation:

\frac{x ^{2}}{4} - \frac{y ^{2}}{b ^{2}} = 1

This hyperbola is centered at the origin

O (0, 0)

and opens along the

x

-axis. Our objective is to determine the area of an equilateral triangle

O P Q

, where

P

and

Q

are points lying on the hyperbola.

Unlocking the Hyperbola

We are given

a^{2} = 4

and the eccentricity

e = 3

. The eccentricity is the fundamental parameter that defines the shape of the conic section.

We utilize the standard relationship for a hyperbola:

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

Substituting the known values into this equation:

3 = 1 + \frac{b ^{2}}{4} \Rightarrow \frac{b ^{2}}{4} = 2 \Rightarrow b^{2} = 8

Thus, the equation of our hyperbola is:

\frac{x ^{2}}{4} - \frac{y ^{2}}{8} = 1

The Geometry of the Chord

Consider the chord

P Q

. Since

Δ O P Q

is equilateral and the hyperbola is symmetric about the

x

-axis, the chord

P Q

must be perpendicular to the

x

-axis.

Let the coordinates of

P

(x_{0}, y_{0})

. Due to symmetry, the coordinates of

Q

are

(x_{0}, - y_{0})

. The length of the chord

P Q

is the vertical distance between these points:

P Q = 2∣ y_{0} ∣

The Algebraic Bridge

For

Δ O P Q

to be equilateral, the distance from the origin to

P

must equal the side length

P Q

. Using the distance formula for

O P

O P = x_{0}^{2} + y_{0}^{2}

Setting

O P = P Q

and squaring both sides, we obtain:

x_{0}^{2} + y_{0}^{2} = (2 y_{0})^{2} \Rightarrow x_{0}^{2} + y_{0}^{2} = 4 y_{0}^{2} \Rightarrow x_{0}^{2} = 3 y_{0}^{2}

Since

P (x_{0}, y_{0})

lies on the hyperbola, it must satisfy the hyperbola equation. Substituting

x_{0}^{2} = 3 y_{0}^{2}

into the equation

\frac{x _{0}^{2}}{4} - \frac{y _{0}^{2}}{8} = 1

\frac{3 y _{0}^{2}}{4} - \frac{y _{0}^{2}}{8} = 1

Multiplying the entire equation by

8

to clear the denominators:

6 y_{0}^{2} - y_{0}^{2} = 8 \Rightarrow 5 y_{0}^{2} = 8 \Rightarrow y_{0}^{2} = \frac{8}{5}

Final Calculation

The area of an equilateral triangle with side length

s

is given by the formula:

Area = \frac{3}{4} s^{2}

In our case, the side length

s = P Q = 2 y_{0}

. Therefore,

s^{2} = 4 y_{0}^{2}

. Substituting this into the area formula:

Area = \frac{3}{4} (4 y_{0}^{2}) = 3 y_{0}^{2}

Substituting the value

y_{0}^{2} = \frac{8}{5}

into the expression:

Area = 3 \times \frac{8}{5} = \frac{8 3}{5}

The area of the equilateral triangle

O P Q

is $\frac{8 3}{5}$ .

Common Traps & FAQs

Wait, why is the side length of the triangle 2y0?

Great question! Since the chord PQ is perpendicular to the x-axis, the points P and Q share the same x-coordinate, say x0. Because the hyperbola is symmetric about the x-axis, if P is at (x0, y0), then Q must be at (x0, -y0). The distance between these two points is the difference in their y-coordinates: y0 - (-y0) = 2y0. This vertical segment is our side length.

Why do we need to find b^2 first?

The equation of the hyperbola is given as x^2/4 - y^2/b^2 = 1. To work with this equation, we need the value of b^2. We are given the eccentricity e = sqrt(3) and a^2 = 4. Using the fundamental relation e^2 = 1 + b^2/a^2, we can solve for b^2. Without this, we cannot define the specific hyperbola we are working with.

Can we use the slope of the line OP to solve this?

You absolutely can! In an equilateral triangle, the angle between the sides is 60 degrees. Since the triangle is symmetric about the x-axis, the line OP makes an angle of 30 degrees with the x-axis. Therefore, the slope of OP is tan(30 degrees) = 1/sqrt(3). This gives you the relation y0/x0 = 1/sqrt(3), or x0 = sqrt(3)y0. Squaring this gives x0^2 = 3y0^2, which is the same relationship we derived using the distance formula. Both methods are perfectly valid and elegant!

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Visualized Solution

Hyperbola Equation and Parameters

Eccentricity Formula

Substituting Known Values

Solving for $b^{2}$

Defining the Chord $P Q$

Equilateral Triangle Condition

Setting Up the Distance Equation

Simplifying the Geometric Condition

Point $P$ on the Hyperbola

Substituting $x_{0}^{2}$

Solving for $y_{0}^{2}$

Area Formula for $Δ O P Q$

Simplifying the Area Expression

Calculating the Final Area

Final Conclusion

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

Analyzing the Setup

Unlocking the Hyperbola

The Geometry of the Chord

The Algebraic Bridge

Final Calculation

Similar Questions

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Comprehension Passage

The orthocentre of the triangle $F_{1} M N$ is

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For $0 < θ < π /2$ , if the eccentricity of the hyperbola $x^{2} - y^{2} cosec^{2} θ = 5$ is $7$ times eccentricity of the ellipse $x^{2} cosec^{2} θ + y^{2} = 5$ , then the value of $θ$ is :

A hyperbola has its centre at the origin, passes through the point (4,2) and has transverse axis of length 4 along the x-axis. Then the eccentricity of the hyperbola is :

Select Answer:

Visualized Solution

Hyperbola Equation and Parameters

Eccentricity Formula

Substituting Known Values

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2

Defining the Chord .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ

Equilateral Triangle Condition

Setting Up the Distance Equation

Simplifying the Geometric Condition

Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P on the Hyperbola

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Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y02​

Area Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔOPQ

Simplifying the Area Expression

Calculating the Final Area

Final Conclusion

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

Analyzing the Setup

Unlocking the Hyperbola

The Geometry of the Chord

The Algebraic Bridge

Final Calculation

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The orthocentre of the triangle F1​MN is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For 0<θ<π/2, if the eccentricity of the hyperbola x2−y2cosec2θ=5 is 7​ times eccentricity of the ellipse x2cosec2θ+y2=5, then the value of θ is :

A hyperbola has its centre at the origin, passes through the point (4,2) and has transverse axis of length 4 along the x-axis. Then the eccentricity of the hyperbola is :

Solving for $b^{2}$

Defining the Chord $P Q$

Point $P$ on the Hyperbola

Substituting $x_{0}^{2}$

Solving for $y_{0}^{2}$

Area Formula for $Δ O P Q$

The orthocentre of the triangle $F_{1} M N$ is

For $0 < θ < π /2$ , if the eccentricity of the hyperbola $x^{2} - y^{2} cosec^{2} θ = 5$ is $7$ times eccentricity of the ellipse $x^{2} cosec^{2} θ + y^{2} = 5$ , then the value of $θ$ is :