Question

Let H: $\frac{-x^2}{a^2} + \frac{y^2}{b^2} = 1$ be the hyperbola, whose eccentricity is $\sqrt{3}$ and the length of the latus rectum is $4\sqrt{3}$. Suppose the point $(\alpha, 6)$, $\alpha>0$ lies on H. If $\beta$ is the product of the focal distances of the point $(\alpha, 6)$, then $\alpha^2 + \beta$ is equal to:

• A. 170
• B. 171
• C. 169
• D. 172

Answer 1

The Correct Option is B

This gives:
\[ a^2 = 2b^2. \]
The length of the latus rectum is given by:
\[ \text{Latus Rectum} = \frac{2a^2}{b}. \]
Substitute \(a^2 = 2b^2\) and Latus Rectum = \(4\sqrt{3}\):
\[ \frac{4b^2}{b} = 4\sqrt{3} \implies 4b = 4\sqrt{3} \implies b = \sqrt{3}. \]
Using \(a^2 = 2b^2\):
\[ a^2 = 2(\sqrt{3})^2 = 2 \cdot 3 = 6 \implies a = \sqrt{6}. \]
The equation of the hyperbola becomes:
\[ \frac{y^2}{3} - \frac{x^2}{6} = 1. \]
The point \((\alpha, 6)\) lies on the hyperbola:
\[ \frac{6^2}{3} - \frac{\alpha^2}{6} = 1 \implies 12 - \frac{\alpha^2}{6} = 1 \implies \frac{\alpha^2}{6} = 11 \implies \alpha^2 = 66. \]
The coordinates of the foci are:
\[ (0, \pm be) = (0, \pm \sqrt{3} \cdot \sqrt{3}) = (0, \pm 3). \]
Let \(d_1\) and \(d_2\) be the focal distances of the point \((\alpha, 6)\):
\[ d_1 = \sqrt{\alpha^2 + (6 - 3)^2}, \quad d_2 = \sqrt{\alpha^2 + (6 + 3)^2}. \]
Substitute:
\[ d_1 = \sqrt{66 + (6 - 3)^2} = \sqrt{66 + 9} = \sqrt{75}, \] \[ d_2 = \sqrt{66 + (6 + 3)^2} = \sqrt{66 + 81} = \sqrt{147}. \]
The product of the focal distances is:
\[ \beta = d_1 \cdot d_2 = \sqrt{75} \cdot \sqrt{147} = \sqrt{75 \cdot 147}. \]
Simplify:
\[ 75 \cdot 147 = 11025 \implies \beta = \sqrt{11025} = 105. \]
Finally, calculate \(a^2 + \beta\):
\[ a^2 + \beta = 66 + 105 = 171. \]
Final Answer: 171.