If the latus rectum of a hyperola forms an equilateral triangle with the centre of the hyperbola, then its eccentricity is

Options

A. `(sqrt5+1)/(2)`
B. `(sqrt11+1)/(2)`
C. `(sqrt13+1)/(2sqrt3)`
D. `(sqrt13*1)/(2sqrt3)`

Answer

To solve the problem step by step, we will analyze the given information about the hyperbola and its latus rectum forming an equilateral triangle with the center.

### Step 1: Understanding the Latus Rectum
The latus rectum of a hyperbola is a line segment perpendicular to the transverse axis and passes through a focus. The ends of the latus rectum can be represented as points on the hyperbola.

### Step 2: Setting Up the Problem
Let the center of the hyperbola be at the origin (0, 0). The ends of the latus rectum can be denoted as (ae, b²/a) and (ae, -b²/a) where 'a' is the semi-major axis, 'b' is the semi-minor axis, and 'e' is the eccentricity.

### Step 3: Equilateral Triangle Condition
Since the latus rectum forms an equilateral triangle with the center, the distance from the center to the ends of the latus rectum must be equal to the length of the latus rectum. The length of the latus rectum (L) for a hyperbola is given by \( L = \frac{2b^2}{a} \).

### Step 4: Using Trigonometric Properties
Given that the angle θ is 30 degrees, we can use the tangent function:
\[ \tan(30^\circ) = \frac{b^2}{ae} \]
Since \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\), we can set up the equation:
\[ \frac{1}{\sqrt{3}} = \frac{b^2}{ae} \]

### Step 5: Relating b² to e
From the equation above, we can express \(b^2\) in terms of \(a\) and \(e\):
\[ b^2 = \frac{ae}{\sqrt{3}} \]

### Step 6: Using the Relationship Between a, b, and e
We know that:
\[ b^2 = e^2 a^2 - a^2 \]
Substituting \(b^2\) from the previous step:
\[ \frac{ae}{\sqrt{3}} = e^2 a^2 - a^2 \]

### Step 7: Rearranging the Equation
Rearranging gives:
\[ e^2 a^2 - a^2 - \frac{ae}{\sqrt{3}} = 0 \]
Factoring out \(a^2\):
\[ a^2(e^2 - 1) = \frac{ae}{\sqrt{3}} \]

### Step 8: Solving for e
Now we can express \(e\) in terms of \(a\):
\[ e^2 - 1 = \frac{e}{\sqrt{3a}} \]
Cross-multiplying leads to a quadratic in \(e\):
\[ \sqrt{3} e^2 - e - \sqrt{3} = 0 \]

### Step 9: Using the Quadratic Formula
Using the quadratic formula \(e = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = \sqrt{3}, b = -1, c = -\sqrt{3}\):
\[ e = \frac{1 \pm \sqrt{1 + 12}}{2\sqrt{3}} = \frac{1 \pm \sqrt{13}}{2\sqrt{3}} \]

### Step 10: Final Result
Thus, the eccentricity \(e\) is:
\[ e = \frac{1 + \sqrt{13}}{2\sqrt{3}} \]

### Conclusion
The eccentricity of the hyperbola is \(\frac{1 + \sqrt{13}}{2\sqrt{3}}\).

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