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

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.