Question

The length of the latus rectum and directrices of a hyperbola with eccentricity $e$ are 9 and $x = \pm \frac{4}{\sqrt{3}}$, respectively. Let the line $y - \sqrt{3}x + \sqrt{3} = 0$ touch this hyperbola at $(x_0, y_0)$. If $m$ is the product of the focal distances of the point $(x_0, y_0)$, then $4e^2 + m$ is equal to ________.

Answer 1

Correct Answer: 61

Given:
\[\frac{2b^2}{a} = 9 \quad \text{and} \quad \frac{a}{e} = \frac{4}{\sqrt{3}}\]
The equation of the tangent \( y - \sqrt{3}x + \sqrt{3} = 0 \) can be rewritten for easier manipulation. The slope \( S \) of this line is:
\[S = \sqrt{3}\]
Using the condition of tangency, we find:
\[6 = 6a^2 - 9\]
\[\implies a = 2, \quad b^2 = 9\]
Thus, the equation of the hyperbola is:
\[\frac{x^2}{4} - \frac{y^2}{9} = 1\]
and for the tangent line:
\[y = \sqrt{3}x + \sqrt{3}\]
The point of contact \( (x_0, y_0) \) is:
\[(4, 3\sqrt{3}) = (x_0, y_0)\]
Now, calculating the eccentricity:
\[e = \sqrt{1 + \frac{9}{4}} = \frac{\sqrt{13}}{2}\]
The product of focal distances \( m \) is given by:
\[m = (x_0 + a)(x_0 - a)\]
\[m + 4e^2 = 20 \times \frac{13}{4} = 61\]
Note: There is a printing mistake in the equation of the directrix as \( x = \pm \frac{4}{\sqrt{3}} \).
Corrected equation: The correct equation for the directrix should be \( x = \pm \frac{4}{\sqrt{13}} \), as eccentricity must be greater than one, so this question might be a bonus.