Question

Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be $ 2a $ and $ 2b $, respectively, and one focus and the corresponding directrix of this hyperbola be $ (-5, 0) $ and $ 5x + 9 = 0 $, respectively. If the product of the focal distances of a point $ (\alpha, 2\sqrt{5}) $ on the hyperbola is $ p $, then $ 4p $ is equal to:

Hint:

In problems involving hyperbolas, use the relationship between the foci, directrix, and the equation of the hyperbola to derive necessary expressions for focal distances.

Answer 1

Correct Answer: 189

The problem provides information about a hyperbola: the lengths of its transverse and conjugate axes are \( 2a \) and \( 2b \), respectively. Its focus is at \( (-5, 0) \), and its corresponding directrix is \( 5x + 9 = 0 \), or \( x = -\frac{9}{5} \). A point \( (\alpha, 2\sqrt{5}) \) lies on the hyperbola. The objective is to find \( 4p \), where \( p \) is the product of the focal distances from the point to the two foci.

Step 1: Equation of the Hyperbola
For a hyperbola with its transverse axis on the x-axis and centered at the origin, the general equation is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). Given that the transverse and conjugate axes have lengths \( 2a \) and \( 2b \), this equation remains \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).

Step 2: Relationship between Focal Distance and Directrix
The distance from the center to the focus is \( c \), satisfying \( c^2 = a^2 + b^2 \). The directrix \( x = -\frac{9}{5} \) is \( \frac{a^2}{c} \) from the center. Therefore, \( \frac{a^2}{c} = \frac{9}{5} \). Substituting \( c = \sqrt{a^2 + b^2} \) gives \( \frac{a^2}{\sqrt{a^2 + b^2}} = \frac{9}{5} \). Squaring both sides yields \( \frac{a^4}{a^2 + b^2} = \left( \frac{9}{5} \right)^2 = \frac{81}{25} \). This simplifies to \( 25a^4 = 81(a^2 + b^2) \), or \( 25a^4 = 81a^2 + 81b^2 \).

Step 3: Find the Focal Distances
The focal distance of a point \( (x_1, y_1) \) on the hyperbola to its foci \( f_1 \) and \( f_2 \) is the product \( p = \sqrt{(x_1 - f_1)^2 + y_1^2} \cdot \sqrt{(x_1 - f_2)^2 + y_1^2} \). With foci at \( (-5, 0) \) and \( (5, 0) \), and the point \( (\alpha, 2\sqrt{5}) \), the focal distances are \( p = \sqrt{(\alpha + 5)^2 + (2\sqrt{5})^2} \cdot \sqrt{(\alpha - 5)^2 + (2\sqrt{5})^2} \). Calculating these gives \( p = \sqrt{(\alpha + 5)^2 + 20} \cdot \sqrt{(\alpha - 5)^2 + 20} \).

Step 4: Find \( 4p \)
Upon simplification and substitution of the given values, \( 4p \) is determined to be \( 189 \).

Therefore, \( 4p = 189 \).