Question

Let \(H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) be a hyperbola such that the distance between its foci is 6 and the distance between its directrices is \(\frac{8}{3}\). If the line \(x = \alpha\) intersects the hyperbola H at the points A and B such that the area of the triangle AOB is \(4\sqrt{15}\), where O is the origin, then \(a^2\) equals:

• A. 12
• B. 16
• C. 24
• D. 25

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Using the distance properties of hyperbolas, determine the parameters \(a\), \(b\), and \(e\) to establish the exact equation of the hyperbola. Finding intersection points of a vertical line \(x = \alpha\) directly models the base and height of the specified triangle AOB.
Step 2: Key Formula or Approach:
Distance between foci \(= 2ae\).
Distance between directrices \(= \frac{2a}{e}\).
Hyperbola relation: \(b^2 = a^2(e^2 - 1)\).
Area of triangle with vertices \((0,0), (\alpha, y), (\alpha, -y)\) is \(\frac{1}{2} \times (2y) \times \alpha = \alpha y\).
Step 3: Detailed Explanation:
Given \(2ae = 6 \implies ae = 3\).
Given \(\frac{2a}{e} = \frac{8}{3} \implies \frac{a}{e} = \frac{4}{3}\).
Multiply them: \((ae)\left(\frac{a}{e}\right) = 3 \times \frac{4}{3} \implies a^2 = 4 \implies a = 2\).
Therefore \(e = \frac{3}{2}\).
Find \(b^2\): \(b^2 = a^2(e^2 - 1) = 4\left(\frac{9}{4} - 1\right) = 5\).
The hyperbola is \(\frac{x^2}{4} - \frac{y^2}{5} = 1\).
Intersect with \(x = \alpha\):
\[ \frac{\alpha^2}{4} - \frac{y^2}{5} = 1 \implies y = \sqrt{5\left(\frac{\alpha^2}{4} - 1\right)} \] The area of \(\Delta AOB\) is \(\alpha y = 4\sqrt{15}\).
Square both sides:
\[ \alpha^2 y^2 = 240 \implies \alpha^2 \left( 5\left(\frac{\alpha^2}{4} - 1\right) \right) = 240 \] \[ \frac{5\alpha^4}{4} - 5\alpha^2 - 240 = 0 \implies \alpha^4 - 4\alpha^2 - 192 = 0 \] Factor the quadratic in \(\alpha^2\):
\[ (\alpha^2 - 16)(\alpha^2 + 12) = 0 \] Since \(\alpha^2\) must be positive, \(\alpha^2 = 16\).
Step 4: Final Answer:
\(\alpha^2\) equals 16.