Let the latus rectum of the hyperbola $\frac{x^2}{9} - \frac{y^2}{b^2} = 1$ subtend an angle of $\frac{\pi}{3}$ at the centre of the hyperbola. If $b^2$ is equal to $\frac{l}{m}(1 + \sqrt{n})$, where $l, m$ and $n$ are co-prime numbers, then $l^2 + m^2 + n^2$ is equal to ______

### Analyzing the Setup We are given the hyperbola defined by the equation: $$\frac{x^2}{9} - \frac{y^2}{b^2} = 1$$ Here, $a^2 = 9$, which implies $a = 3$. The latus rectum is a vertical chord passing through the focus $F(ae, 0)$. The endpoints of the latus rectum are $P(ae, \frac{b^2}{a})$ and $Q(ae, -\frac{b^2}{a})$. We are given that this segment subtends an angle of $\frac{\pi}{3}$ at the center $(0,0)$. ### Visualizing the Geometry Due to the symmetry of the hyperbola about the $x$-axis, the line segment from the origin to the point $P$ makes an angle of $\frac{\pi}{6}$ with the $x$-axis. Consider the right-angled triangle formed by the origin $(0,0)$, the focus $(ae, 0)$, and the point $P(ae, \frac{b^2}{a})$. The tangent of the angle $\frac{\pi}{6}$ is the ratio of the vertical height to the horizontal distance: $$\tan\left(\frac{\pi}{6}\right) = \frac{\frac{b^2}{a}}{ae} = \frac{b^2}{a^2e}$$ Substituting $a^2 = 9$ into this expression, we obtain: $$\frac{1}{\sqrt{3}} = \frac{b^2}{9e} \implies e = \frac{\sqrt{3}b^2}{9}$$ ### The Algebraic Dance We utilize the fundamental eccentricity identity for a hyperbola, $e^2 = 1 + \frac{b^2}{a^2}$. Substituting $a^2 = 9$, we have: $$e^2 = 1 + \frac{b^2}{9}$$ Squaring our previous expression for $e$, we get: $$e^2 = \left( \frac{\sqrt{3}b^2}{9} \right)^2 = \frac{3b^4}{81} = \frac{b^4}{27}$$ Equating the two expressions for $e^2$: $$\frac{b^4}{27} = 1 + \frac{b^2}{9}$$ Multiplying the entire equation by $27$ yields the quadratic form: $$b^4 - 3b^2 - 27 = 0$$ ### Final Calculation Applying the quadratic formula to solve for $b^2$: $$b^2 = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-27)}}{2} = \frac{3 \pm \sqrt{9 + 108}}{2} = \frac{3 \pm \sqrt{117}}{2}$$ Since $b^2$ must be positive, we take the positive root: $$b^2 = \frac{3 + 3\sqrt{13}}{2} = \frac{3}{2}(1 + \sqrt{13})$$ Comparing this to the form $\frac{l}{m}(1 + \sqrt{n})$, we identify $l=3$, $m=2$, and $n=13$. These values are co-prime. The final result is: $$l^2 + m^2 + n^2 = 3^2 + 2^2 + 13^2 = 9 + 4 + 169 = \mathbf{182}$$

JEE Main 2024 (30 Jan Shift 1)

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Animated Solution for Mathematics - Conic Sections: Let the latus rectum of the hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{b ^{2}} = 1$ subtend an angle of $\frac{π}{3}$ at the centre of the hyperbola. If $b^{2}$ is equal to $\frac{l}{m} (1 + n)$ , where $l, m$ and $n$ are co-prime numbers, then $l^{2} + m^{2} + n^{2}$ is equal to ______

Q: Why do we reject the negative root when solving the quadratic equation for $b^2$?

In the standard equation of a hyperbola, $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the term $b^2$ represents the square of a real number. By definition, $b^2$ must be a positive value. If we accepted the negative root, we would be implying that $b^2$ is negative, which would violate the geometric reality of the hyperbola. Always keep the physical constraints of the conic section in mind!

Q: Can we solve this without using the eccentricity identity $e^2 = 1 + \frac{b^2}{a^2}$?

You could try to express everything in terms of $a, b,$ and $e$ using other relations, but the identity $e^2 = 1 + \frac{b^2}{a^2}$ is the most direct bridge between the geometric parameters and the algebraic constraints. In JEE problems, recognizing these standard identities is key to saving time and avoiding messy, unnecessary calculations.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the geometry. The latus rectum is a vertical line segment passing through the focus at

(a e, 0)

. The center of the hyperbola is at the origin

(0, 0)

. The angle subtended is at the origin. Consider the right-angled triangle formed by the center, the focus, and the upper endpoint of the latus rectum,

P (a e, \frac{b ^{2}}{a})

Hint 2

Hint 2 (The Arsenal): Use the definition of the tangent function in the right-angled triangle.

tan (θ) = \frac{opposite}{adjacent}

. Here, the opposite side is the y-coordinate of the latus rectum endpoint, and the adjacent side is the x-coordinate (the focus). Remember that the total angle is

\frac{π}{3}

, so the angle

θ

in your triangle is

\frac{π}{6}

Hint 3

Hint 3 (The Execution): Start with the equation

tan (3 0^{\circ}) = \frac{b ^{2} / a}{a e} = \frac{b ^{2}}{a ^{2} e}

. Substitute

a^{2} = 9

and use the eccentricity identity

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

to eliminate

e

and solve for

b^{2}

Visualized Solution

Identify the Hyperbola Equation

Given Hyperbola: $\frac{x ^{2}}{9} - \frac{y ^{2}}{b ^{2}} = 1$
Comparing with standard form $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ : $a^{2} = 9 ⟹ a = 3$

Locate the Latus Rectum

Endpoints of Latus Rectum: $P (a e, \frac{b ^{2}}{a})$ and $Q (a e, - \frac{b ^{2}}{a})$
Focus is at $(a e, 0)$

Analyze the Subtended Angle

Total angle at center $C (0, 0) = \frac{π}{3} = 6 0^{\circ}$
By symmetry, half angle $θ = \frac{π}{6} = 3 0^{\circ}$

Apply Trigonometry

In right $△ C F P$ (where $F$ is focus): $tan (3 0^{\circ}) = \frac{Perpendicular}{Base}$
$tan (3 0^{\circ}) = \frac{\frac{b ^{2}}{a}}{a e} = \frac{b ^{2}}{a ^{2} e}$

Substitute Known Values

Substitute $a^{2} = 9$ and $tan (3 0^{\circ}) = \frac{1}{3}$
$\frac{1}{3} = \frac{b ^{2}}{9 e}$

Express Eccentricity $e$

Rearranging for $e$ : $e = \frac{3 b ^{2}}{9}$

Use the Eccentricity Identity

Standard Identity for Hyperbola: $e^{2} = 1 + \frac{b ^{2}}{a ^{2}}$
Substituting $a^{2} = 9$ : $e^{2} = 1 + \frac{b ^{2}}{9}$

Substitute $e$ into Identity

Substitute $e = \frac{3 b ^{2}}{9}$ :
$(\frac{3 b ^{2}}{9})^{2} = 1 + \frac{b ^{2}}{9}$

Simplify the Equation

Expand the square: $\frac{3 b ^{4}}{81} = \frac{9 + b ^{2}}{9}$
Simplify fraction: $\frac{b ^{4}}{27} = \frac{9 + b ^{2}}{9}$

Form Quadratic Equation

Cross-multiply: $b^{4} = 3 (9 + b^{2})$
Rearrange: $b^{4} - 3 b^{2} - 27 = 0$

Solve the Quadratic Equation

Let $x = b^{2}$ . Equation is $x^{2} - 3 x - 27 = 0$
Using Quadratic Formula: $b^{2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( 1 ) ( - 27 )}{2 ( 1 )}$

Simplify the Radical

$b^{2} = \frac{3 \pm 9 + 108}{2} = \frac{3 \pm 117}{2}$
Since $b^{2} > 0$ , take positive root: $b^{2} = \frac{3 + 117}{2}$
$117 = 9 \times 13 = 313$
$b^{2} = \frac{3}{2} (1 + 13)$

Compare and Find Constants

Given form: $b^{2} = \frac{l}{m} (1 + n)$
Comparing: $l = 3$ , $m = 2$ , $n = 13$
Check: $gcd (3, 2, 13) = 1$ (They are co-prime)

Final Calculation

Calculate $l^{2} + m^{2} + n^{2}$
$3^{2} + 2^{2} + 1 3^{2} = 9 + 4 + 169$
Final Result: $182$

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

Solution Diagram

Analyzing the Setup

We are given the hyperbola defined by the equation:

\frac{x ^{2}}{9} - \frac{y ^{2}}{b ^{2}} = 1

Here,

a^{2} = 9

, which implies

a = 3

. The latus rectum is a vertical chord passing through the focus

F (a e, 0)

The endpoints of the latus rectum are

P (a e, \frac{b ^{2}}{a})

and

Q (a e, - \frac{b ^{2}}{a})

. We are given that this segment subtends an angle of

\frac{π}{3}

at the center

(0, 0)

Visualizing the Geometry

Due to the symmetry of the hyperbola about the

x

-axis, the line segment from the origin to the point

P

makes an angle of

\frac{π}{6}

with the

x

-axis.

Consider the right-angled triangle formed by the origin

(0, 0)

, the focus

(a e, 0)

, and the point

P (a e, \frac{b ^{2}}{a})

. The tangent of the angle

\frac{π}{6}

is the ratio of the vertical height to the horizontal distance:

tan (\frac{π}{6}) = \frac{\frac{b ^{2}}{a}}{a e} = \frac{b ^{2}}{a ^{2} e}

Substituting

a^{2} = 9

into this expression, we obtain:

\frac{1}{3} = \frac{b ^{2}}{9 e} ⟹ e = \frac{3 b ^{2}}{9}

The Algebraic Dance

We utilize the fundamental eccentricity identity for a hyperbola,

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

. Substituting

a^{2} = 9

, we have:

e^{2} = 1 + \frac{b ^{2}}{9}

Squaring our previous expression for

e

, we get:

e^{2} = (\frac{3 b ^{2}}{9})^{2} = \frac{3 b ^{4}}{81} = \frac{b ^{4}}{27}

Equating the two expressions for

e^{2}

\frac{b ^{4}}{27} = 1 + \frac{b ^{2}}{9}

Multiplying the entire equation by

27

yields the quadratic form:

b^{4} - 3 b^{2} - 27 = 0

Final Calculation

Applying the quadratic formula to solve for

b^{2}

b^{2} = \frac{3 \pm ( - 3 ) ^{2} - 4 ( 1 ) ( - 27 )}{2} = \frac{3 \pm 9 + 108}{2} = \frac{3 \pm 117}{2}

Since

b^{2}

must be positive, we take the positive root:

b^{2} = \frac{3 + 3 13}{2} = \frac{3}{2} (1 + 13)

Comparing this to the form

\frac{l}{m} (1 + n)

, we identify

l = 3

m = 2

, and

n = 13

. These values are co-prime.

The final result is:

l^{2} + m^{2} + n^{2} = 3^{2} + 2^{2} + 1 3^{2} = 9 + 4 + 169 = 182

Common Traps & FAQs

Wait, why do we use

tan (3 0^{\circ})

and not

tan (6 0^{\circ})

That is a fantastic observation! The problem states the *entire* latus rectum subtends an angle of

\frac{π}{3}

(

6 0^{\circ}

) at the center. However, the hyperbola is perfectly symmetric about the x-axis. The line from the center to the focus lies on the x-axis, which bisects that

6 0^{\circ}

angle. Therefore, the triangle we are analyzing only covers half of that angle, which is

\frac{6 0 ^{\circ}}{2} = 3 0^{\circ}

Why do we reject the negative root when solving the quadratic equation for

b^{2}

In the standard equation of a hyperbola,

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

, the term

b^{2}

represents the square of a real number. By definition,

b^{2}

must be a positive value. If we accepted the negative root, we would be implying that

b^{2}

is negative, which would violate the geometric reality of the hyperbola. Always keep the physical constraints of the conic section in mind!

Can we solve this without using the eccentricity identity

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

You could try to express everything in terms of

a, b,

and

e

using other relations, but the identity

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

is the most direct bridge between the geometric parameters and the algebraic constraints. In JEE problems, recognizing these standard identities is key to saving time and avoiding messy, unnecessary calculations.

Enter Numerical Value:

Visualized Solution

Identify the Hyperbola Equation

Locate the Latus Rectum

Analyze the Subtended Angle

Apply Trigonometry

Substitute Known Values

Express Eccentricity $e$

Use the Eccentricity Identity

Substitute $e$ into Identity

Simplify the Equation

Form Quadratic Equation

Solve the Quadratic Equation

Simplify the Radical

Compare and Find Constants

Final Calculation

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

Analyzing the Setup

Visualizing the Geometry

The Algebraic Dance

Final Calculation

Similar Questions

Consider the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ having one of its focus at $P (- 3, 0)$ . If the latus ractum through its other focus subtends a right angle at $P$ and $a^{2} b^{2} = α 2 - β, α, β \in N$ , then $α + β$ is equal to

Let the eccentricity of an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ is reciprocal to that of the hyperbola $2 x^{2} - 2 y^{2} = 1$ . If the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is _____.

Let the foci of the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1$ and the hyperbola $\frac{x ^{2}}{144} - \frac{y ^{2}}{α} = \frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is:-

Let the foci of a hyperbola coincide with the foci of the ellipse $\frac{x ^{2}}{36} + \frac{y ^{2}}{16} = 1$ . If the eccentricity of the hyperbola is 5, then the length of its latus rectum is :

Enter Numerical Value:

Visualized Solution

Identify the Hyperbola Equation

Locate the Latus Rectum

Analyze the Subtended Angle

Apply Trigonometry

Substitute Known Values

Express Eccentricity .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e

Use the Eccentricity Identity

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e into Identity

Simplify the Equation

Form Quadratic Equation

Solve the Quadratic Equation

Simplify the Radical

Compare and Find Constants

Final Calculation

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

Analyzing the Setup

Visualizing the Geometry

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 the foci of the ellipse 16x2​+7y2​=1 and the hyperbola 144x2​−αy2​=251​ coincide. Then the length of the latus rectum of the hyperbola is:-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the foci of a hyperbola coincide with the foci of the ellipse 36x2​+16y2​=1. If the eccentricity of the hyperbola is 5, then the length of its latus rectum is :

Express Eccentricity $e$

Substitute $e$ into Identity

Let the foci of the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1$ and the hyperbola $\frac{x ^{2}}{144} - \frac{y ^{2}}{α} = \frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is:-

Let the foci of a hyperbola coincide with the foci of the ellipse $\frac{x ^{2}}{36} + \frac{y ^{2}}{16} = 1$ . If the eccentricity of the hyperbola is 5, then the length of its latus rectum is :