Let H: x²/a² - y²/b² = 1, where a b 0, be a hyperbola in the xy-plane whose conjugate axis LM su

Q: Wait, why is the angle $\angle ONL = 30^\circ$ and not $60^\circ$?

Great question! The hyperbola $H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is symmetric about both the $x$-axis and the $y$-axis. The conjugate axis $LM$ is a vertical line segment centered at the origin. The vertex $N$ lies on the $x$-axis. Because the entire figure is symmetric about the $x$-axis, the $x$-axis acts as an angle bisector for $\angle LNM$. Since the total angle is $60^\circ$, the portion above the $x$-axis, which is $\angle ONL$, must be exactly half of that, which is $30^\circ$.

Q: Why can't we just use the definition of the hyperbola $x^2/a^2 - y^2/b^2 = 1$ to find $a$ and $b$ directly?

The definition of the hyperbola gives you the *shape*, but it doesn't give you the specific values of $a$ and $b$ without additional constraints. In this problem, the geometric constraints (the subtended angle and the area) are the keys that unlock the specific values of $a$ and $b$. Once you have those, the standard equation becomes fully defined, and you can calculate all the properties like eccentricity and latus rectum.

Q: Is the area of triangle $LMN$ always $ab$?

Let's look at the geometry. The triangle $LMN$ has a base $LM$ which is the conjugate axis, so its length is $2b$. The height of this triangle, relative to the base $LM$, is the distance from the origin to the vertex $N$, which is $a$. The area of any triangle is $\frac{1}{2} \times \text{base} \times \text{height}$. So, $\text{Area} = \frac{1}{2} \times (2b) \times a = ab$. Yes, it is always $ab$ for this specific configuration!

Question

Let $H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, where $a > b > 0$, be a hyperbola in the $xy$-plane whose conjugate axis $LM$ subtends an angle of $60^\circ$ at one of its vertices $N$. Let the area of the triangle $LMN$ be $4\sqrt{3}$. Match the properties in List-I to the numbers in List-II.

List-I:

P. The length of the conjugate axis of $H$ is

Q. The eccentricity of $H$ is

R. The distance between the foci of $H$ is

S. The length of the latus rectum of $H$ is

List-II:

1. 8

2. 4

3. $\frac{2}{\sqrt{3}}$

4. $4\sqrt{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are peeling back the layers of a beautiful geometric structure: the hyperbola.

When you look at the equation $H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, do not see it as a cold, static arrangement of variables. See it as a dance of symmetry and proportion. Our goal is to find the parameters $a$ and $b$ that define this specific hyperbola, and from there, unveil its hidden properties.

### Visualizing the Geometry

Imagine you are standing on the $xy$-plane. You have a hyperbola centered at the origin. Its vertices are at $(\pm a, 0)$, and its conjugate axis endpoints, $L$ and $M$, are at $(0, b)$ and $(0, -b)$.

The problem introduces a fascinating constraint: the conjugate axis $LM$ subtends an angle of $60^{\circ}$ at one of the vertices, $N(a, 0)$.

Draw this. Sketch the hyperbola, mark the points $L(0, b)$, $M(0, -b)$, and $N(a, 0)$. Connect $L$ to $N$ and $M$ to $N$. You have created a triangle $	riangle LMN$ where $\angle LNM = 60^{\circ}$.

### The Symmetry Insight

Mathematics is often about finding the right perspective. Because our hyperbola is perfectly symmetric about the $x$-axis, the $x$-axis acts as a mirror. It cuts the triangle $	riangle LMN$ in half.

This means the $x$-axis bisects the angle $\angle LNM$. If the total angle is $60^{\circ}$, then the angle in the upper right-angled triangle $	riangle ONL$, which is $\angle ONL$, must be exactly $30^{\circ}$.

We have turned a complex triangle problem into a simple trigonometry problem in a right-angled triangle.

### The Algebraic Bridge

Now, let's use our toolkit. In the right-angled triangle $	riangle ONL$, we have:

$$	an(\angle ONL) = \frac{OL}{ON}$$

We know $OL = b$ and $ON = a$. So, $	an(30^{\circ}) = \frac{b}{a}$. Since $	an(30^{\circ}) = \frac{1}{\sqrt{3}}$, we arrive at the elegant relation:

$$a = b\sqrt{3}$$

This is our first bridge between $a$ and $b$. We now require the area of $	riangle LMN$ to solve for them individually.

### The Area Calculation

The area of $	riangle LMN$ is given as $4\sqrt{3}$. The base of this triangle is the conjugate axis $LM$, which has length $2b$. The height of the triangle, from the origin to the vertex $N$, is $a$.

Therefore, the area is:

$$	ext{Area} = \frac{1}{2} 	imes (2b) 	imes a = ab$$

We are given $	ext{Area} = 4\sqrt{3}$, so we have our second equation:

$$ab = 4\sqrt{3}$$

### Solving for $a$ and $b$

Now, we have a system of two equations:
1) $a = b\sqrt{3}$
2) $ab = 4\sqrt{3}$

Substitute the first into the second:

$$(b\sqrt{3})b = 4\sqrt{3}$$

$$b^2\sqrt{3} = 4\sqrt{3}$$

Canceling $\sqrt{3}$ from both sides, we get $b^2 = 4$. Since $b > 0$, we find **$b = 2$**. Substituting this back into our first equation, we get **$a = 2\sqrt{3}$**.

### Unveiling the Properties

With $a = 2\sqrt{3}$ and $b = 2$, the hyperbola is fully revealed. Now we can calculate the properties requested:

1. **Length of the conjugate axis:** This is $2b = 2(2) = 4$.

2. **Eccentricity ($e$):** The formula is $e = \sqrt{1 + \frac{b^2}{a^2}}$. Substituting our values:

$$e = \sqrt{1 + \frac{4}{12}} = \sqrt{1 + \frac{1}{3}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$$

3. **Distance between the foci:** The foci are at $(\pm ae, 0)$, so the distance is:

$$2ae = 2(2\sqrt{3})\left(\frac{2}{\sqrt{3}}ight) = 8$$

4. **Length of the latus rectum:** The formula is:

$$\frac{2b^2}{a} = \frac{2(4)}{2\sqrt{3}} = \frac{4}{\sqrt{3}}$$

And there you have it! By simply visualizing the geometry and applying the symmetry of the hyperbola, we have unlocked all its secrets. Remember, every complex problem is just a collection of simple, elegant truths waiting to be connected.

List-I

List-II

Select Matching Pairs:

Visualized Solution

Standard Hyperbola Setup

Subtended Angle at Vertex

Symmetry and Half-Angle

Trigonometry in $△ O N L$

Area of Triangle $L M N$

Solving for $a$ and $b$

Property P: Conjugate Axis Length

Property Q: Eccentricity

Property R: Distance Between Foci

Property S: Latus Rectum

The Sigma Insight: Foci, Directrices, and Eccentricity

Analyzing the Setup

Visualizing the Geometry

The Symmetry Insight

The Algebraic Bridge

The Area Calculation

Solving for $a$ and $b$

Unveiling the Properties

Similar Questions

The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is :

Let H be the hyperbola, whose foci are $(1 \pm 2, 0)$ and eccentricity is $2$ . Then the length of its latus rectum is

Let the product of the focal distances of the point $P (4, 23)$ on the hyperbola $H : \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ be 32. Let the length of the conjugate axis of $H$ be $p$ and the length of its latus rectum be $q$ . Then $p^{2} + q^{2}$ is equal to _______

Let $0 < θ < \frac{π}{2}$ . If the eccentricity of the hyperbola $\frac{x ^{2}}{c o s ^{2} θ} - \frac{y ^{2}}{s i n ^{2} θ} = 1$ is greater than 2, then the length of its latus rectum lies in the interval :

Let $H : \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1, a > 0, b > 0$ , be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is $4 (22 + 14)$ . If the eccentricity $H$ is $\frac{11}{2}$ , then value of $a^{2} + b^{2}$ is equal to ______.

Let one focus of the hyperbola $H : \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ be at $(10, 0)$ and the corresponding directrix be $x = \frac{9}{10}$ . If $e$ and $l$ respectively are the eccentricity and the length of the latus rectum of $H$ , then $9 (e^{2} + l)$ is equal to:

Let an ellipse with centre $(1, 0)$ and latus rectum of length $\frac{1}{2}$ have its major axis along x-axis. If its minor axis subtends an angle $6 0^{\circ}$ at the foci, then the square of the sum of the lengths of its minor and major axes is equal to _______.

If a hyperbola has length of its conjugate axis equal to 5 and the distance between its foci is 13, then the eccentricity of the hyperbola is:

Let the eccentricity of the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ be a reciprocal to that of the ellipse $x^{2} + 4 y^{2} = 4$ . If the hyperbola passes through a focus of the ellipse, then

List-I

List-II

Select Matching Pairs:

Visualized Solution

Standard Hyperbola Setup

Subtended Angle at Vertex

Symmetry and Half-Angle

Trigonometry in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △ONL

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

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

Property P: Conjugate Axis Length

Property Q: Eccentricity

Property R: Distance Between Foci

Property S: Latus Rectum

The Sigma Insight: Foci, Directrices, and Eccentricity

Analyzing the Setup

Visualizing the Geometry

The Symmetry Insight

The Algebraic Bridge

The Area Calculation

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

Unveiling the Properties

Similar Questions

The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let H be the hyperbola, whose foci are (1±2​,0) and eccentricity is 2​. Then the length of its latus rectum is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let 0<θ<2π​. If the eccentricity of the hyperbola cos2θx2​−sin2θy2​=1 is greater than 2, then the length of its latus rectum lies in the interval :

If a hyperbola has length of its conjugate axis equal to 5 and the distance between its foci is 13, then the eccentricity of the hyperbola is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the eccentricity of the hyperbola a2x2​−b2y2​=1 be a reciprocal to that of the ellipse x2+4y2=4. If the hyperbola passes through a focus of the ellipse, then

Trigonometry in $△ O N L$

Area of Triangle $L M N$

Solving for $a$ and $b$

Solving for $a$ and $b$

Let H be the hyperbola, whose foci are $(1 \pm 2, 0)$ and eccentricity is $2$ . Then the length of its latus rectum is

Let $0 < θ < \frac{π}{2}$ . If the eccentricity of the hyperbola $\frac{x ^{2}}{c o s ^{2} θ} - \frac{y ^{2}}{s i n ^{2} θ} = 1$ is greater than 2, then the length of its latus rectum lies in the interval :

Let the eccentricity of the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ be a reciprocal to that of the ellipse $x^{2} + 4 y^{2} = 4$ . If the hyperbola passes through a focus of the ellipse, then