Let `H :(x^2)/(a^2)-(y^2)/(b^2)=1` , where `a > b >0` , be a hyperbola in the `x y` -plane whose conjugate axis `L M` subtends an angle of `60o` at one of its vertices `N` . Let the area of the triangle `L M N` be `4sqrt(3)` . LIST-I LIST-II P. The length of the conjugate axis of `H` is 1. `8` Q. The eccentricity of `H` is 2. `(sqrt(4))/3` R. The distance between the foci of `H` is 3. `2/(sqrt(3))` S. The length of the latus rectum of `H` is 4. `4` The correct option is `P->4;\ \ Q->2;\ \ R->1;\ \ S->3` (b) `P->4;\ \ Q->3;\ \ R->1;\ \ S->2` (c) `P->4;\ \ Q->1;\ \ R->3;\ \ S->2` (d) `P->3;\ \ Q->4;\ \ R->2;\ \ S->1`

Options

A. (P-4,Q-2,R-1,S-3)
B. (P-4,Q-3,R-1,S-2)
C. (P-4,Q-1,R-3,S-2)
D. (P-3,Q-4,R-2,S-1)

Answer

To solve the problem, we need to find the values of the conjugate axis length, eccentricity, distance between the foci, and the length of the latus rectum of the hyperbola given by the equation:

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

where \( a > b > 0 \).

### Step 1: Understanding the Geometry
The conjugate axis \( LM \) subtends an angle of \( 60^\circ \) at one of the vertices \( N \). This means that the angle between the lines connecting the vertex to the endpoints of the conjugate axis is \( 60^\circ \).

### Step 2: Area of Triangle \( LMN \)
The area of triangle \( LMN \) is given as \( 4\sqrt{3} \). The area of a triangle can be expressed as:

\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]

Here, the base \( LM \) is equal to \( 2b \) (the length of the conjugate axis), and the height is \( a \) (the distance from the vertex to the line containing \( LM \)). Therefore, we have:

\[ 4\sqrt{3} = \frac{1}{2} \times 2b \times a \]

This simplifies to:

\[ 4\sqrt{3} = b \times a \]

### Step 3: Finding \( b \) in terms of \( a \)
From the geometry of the hyperbola and the angle, we can use the tangent of \( 30^\circ \):

\[ \tan(30^\circ) = \frac{b}{a} \implies b = a \cdot \tan(30^\circ) = a \cdot \frac{1}{\sqrt{3}} = \frac{a}{\sqrt{3}} \]

### Step 4: Substitute \( b \) into the Area Equation
Substituting \( b \) back into the area equation:

\[ 4\sqrt{3} = \left(\frac{a}{\sqrt{3}}\right) \cdot a \]

This gives:

\[ 4\sqrt{3} = \frac{a^2}{\sqrt{3}} \]

Multiplying both sides by \( \sqrt{3} \):

\[ 12 = a^2 \implies a = \sqrt{12} = 2\sqrt{3} \]

### Step 5: Finding \( b \)
Now substituting \( a \) back to find \( b \):

\[ b = \frac{a}{\sqrt{3}} = \frac{2\sqrt{3}}{\sqrt{3}} = 2 \]

### Step 6: Length of the Conjugate Axis
The length of the conjugate axis is given by:

\[ \text{Length of conjugate axis} = 2b = 2 \times 2 = 4 \]

### Step 7: Eccentricity of the Hyperbola
The eccentricity \( e \) of the hyperbola is given by:

\[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{2^2}{(2\sqrt{3})^2}} = \sqrt{1 + \frac{4}{12}} = \sqrt{1 + \frac{1}{3}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}} \]

### Step 8: Distance Between the Foci
The distance between the foci is given by:

\[ 2ae = 2 \cdot 2\sqrt{3} \cdot \frac{2}{\sqrt{3}} = 8 \]

### Step 9: Length of the Latus Rectum
The length of the latus rectum \( L \) is given by:

\[ L = \frac{2b^2}{a} = \frac{2 \cdot 2^2}{2\sqrt{3}} = \frac{8}{2\sqrt{3}} = \frac{4}{\sqrt{3}} \]

### Summary of Results
- Length of the conjugate axis \( P = 4 \)
- Eccentricity \( Q = \frac{2}{\sqrt{3}} \)
- Distance between the foci \( R = 8 \)
- Length of the latus rectum \( S = \frac{4}{\sqrt{3}} \)

### Matching with Options
From the results:
- \( P \to 4 \)
- \( Q \to 2 \) (as \( \frac{2}{\sqrt{3}} \))
- \( R \to 1 \) (as \( 8 \))
- \( S \to 3 \) (as \( \frac{4}{\sqrt{3}} \))

Thus, the correct option is:
**(b) \( P \to 4; Q \to 3; R \to 1; S \to 2 \)**