If the extremities of the latus rectum of the hyperbola with positive coordinates lie on the parabola `x^(2) = 3(y + 3)`, then length of the latus rectum of the hyperbola when its eccentricity is `sqrt(3)` is

Options

A. 3
B. 6
C. 12
D. none of these

Answer

To find the length of the latus rectum of the hyperbola given that its extremities lie on the parabola \(x^2 = 3(y + 3)\) and the eccentricity is \(\sqrt{3}\), we can follow these steps:

### Step 1: Identify the Hyperbola
Assume the hyperbola is of the form:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
where \(a\) and \(b\) are the semi-major and semi-minor axes respectively.

### Step 2: Find the Length of the Latus Rectum
The length of the latus rectum \(L\) of the hyperbola is given by:
\[ L = \frac{2b^2}{a} \]

### Step 3: Use the Eccentricity
The eccentricity \(e\) of the hyperbola is related to \(a\) and \(b\) by the formula:
\[ e^2 = 1 + \frac{b^2}{a^2} \]
Given that \(e = \sqrt{3}\), we have:
\[ 3 = 1 + \frac{b^2}{a^2} \]
This simplifies to:
\[ \frac{b^2}{a^2} = 2 \quad \Rightarrow \quad b^2 = 2a^2 \]

### Step 4: Substitute \(b^2\) into the Length of Latus Rectum
Substituting \(b^2 = 2a^2\) into the formula for the length of the latus rectum:
\[ L = \frac{2(2a^2)}{a} = \frac{4a^2}{a} = 4a \]

### Step 5: Find the Coordinates of the Extremities of the Latus Rectum
The extremities of the latus rectum are located at:
\[ \left(a, \frac{b^2}{a}\right) \quad \text{and} \quad \left(-a, \frac{b^2}{a}\right) \]
Substituting \(b^2 = 2a^2\):
\[ \left(a, \frac{2a^2}{a}\right) = \left(a, 2a\right) \]

### Step 6: Ensure Points Lie on the Parabola
The points \((a, 2a)\) must satisfy the equation of the parabola:
\[ x^2 = 3(y + 3) \]
Substituting \(x = a\) and \(y = 2a\):
\[ a^2 = 3(2a + 3) \]
This simplifies to:
\[ a^2 = 6a + 9 \]
Rearranging gives:
\[ a^2 - 6a - 9 = 0 \]

### Step 7: Solve the Quadratic Equation
Using the quadratic formula \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ a = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-9)}}{2 \cdot 1} = \frac{6 \pm \sqrt{36 + 36}}{2} = \frac{6 \pm \sqrt{72}}{2} = \frac{6 \pm 6\sqrt{2}}{2} = 3 \pm 3\sqrt{2} \]
Since we need positive coordinates, we take:
\[ a = 3 + 3\sqrt{2} \]

### Step 8: Calculate \(b^2\)
Using \(b^2 = 2a^2\):
\[ b^2 = 2(3 + 3\sqrt{2})^2 = 2(9 + 18\sqrt{2} + 18) = 2(27 + 18\sqrt{2}) = 54 + 36\sqrt{2} \]

### Step 9: Find the Length of the Latus Rectum
Substituting \(a\) back into the length formula:
\[ L = 4a = 4(3 + 3\sqrt{2}) = 12 + 12\sqrt{2} \]

### Final Answer
Thus, the length of the latus rectum of the hyperbola is:
\[ \boxed{12} \]