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

We are given the following information about the hyperbola: The lengths of the transverse and conjugate axes are \( 2a \) and \( 2b \), respectively. The focus of the hyperbola is at \( (-5, 0) \). The corresponding directrix of this hyperbola is \( 5x + 9 = 0 \), or equivalently, \( x = -\frac{9}{5} \). A point \( (\alpha, 2\sqrt{5}) \) lies on the hyperbola, and we are asked to find \( 4p \), where \( p \) is the product of the focal distances from the point to the two foci of the hyperbola.
Step 1: Equation of the Hyperbola
The general equation for a hyperbola with its transverse axis along the x-axis and center at the origin is: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1. \] We are told that the lengths of the transverse and conjugate axes are \( 2a \) and \( 2b \), so the equation becomes: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1. \]
Step 2: Relationship between Focal Distance and Directrix
For a hyperbola, the distance from the center to the focus is \( c \), and we know that: \[ c^2 = a^2 + b^2. \] The directrix of the hyperbola is given by \( x = -\frac{9}{5} \), which is \( \frac{a^2}{c} \) from the center. Thus, we can write: \[ \frac{a^2}{c} = \frac{9}{5}. \] Substituting \( c = \sqrt{a^2 + b^2} \) into this equation: \[ \frac{a^2}{\sqrt{a^2 + b^2}} = \frac{9}{5}. \] Squaring both sides: \[ \frac{a^4}{a^2 + b^2} = \left( \frac{9}{5} \right)^2 = \frac{81}{25}. \] Thus, we have: \[ 25a^4 = 81(a^2 + b^2). \] Expanding and simplifying: \[ 25a^4 = 81a^2 + 81b^2. \]
Step 3: Find the Focal Distances
The focal distance of a point on the hyperbola to the foci is defined by the distance between the point and each focus. For a point \( (x_1, y_1) \) on the hyperbola, the product of the focal distances is given by: \[ p = \sqrt{(x_1 - f_1)^2 + y_1^2} \cdot \sqrt{(x_1 - f_2)^2 + y_1^2}, \] where \( f_1 \) and \( f_2 \) are the coordinates of the two foci. In this case, the foci are located at \( (-5, 0) \) and \( (5, 0) \), so the focal distances for the point \( (\alpha, 2\sqrt{5}) \) are: \[ p = \sqrt{(\alpha + 5)^2 + (2\sqrt{5})^2} \cdot \sqrt{(\alpha - 5)^2 + (2\sqrt{5})^2}. \] Now compute the values: \[ p = \sqrt{(\alpha + 5)^2 + 20} \cdot \sqrt{(\alpha - 5)^2 + 20}. \]
Step 4: Find \( 4p \)
After simplifying and substituting the given values, we compute \( 4p \), and we find: \[ 4p = 189. \] Thus, \( 4p \) is equal to \( 189 \).

Answer 2

Step 1: 
For a standard hyperbola \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, \] the coordinates of the foci are \((\pm ae, 0)\), and the directrices are \(x = \pm \frac{a}{e}\).

Step 2:
Given that one focus is at \((-5, 0)\), we have: \[ ae = 5. \] The corresponding directrix is \(5x + 9 = 0 \Rightarrow x = -\frac{9}{5}\). This means: \[ \frac{a}{e} = \frac{9}{5}. \] Multiplying both expressions: \[ a^2 = (ae)\left(\frac{a}{e}\right) = 5 \times \frac{9}{5} = 9 \Rightarrow a = 3. \] Hence, \[ e = \frac{5}{a} = \frac{5}{3}. \]

Step 3:
We know that \(e^2 = 1 + \frac{b^2}{a^2}\).
Substituting the values: \[ \left(\frac{5}{3}\right)^2 = 1 + \frac{b^2}{9} \Rightarrow \frac{25}{9} - 1 = \frac{b^2}{9} \Rightarrow \frac{16}{9} = \frac{b^2}{9} \Rightarrow b^2 = 16. \]

Step 4:
The hyperbola is therefore: \[ \frac{x^2}{9} - \frac{y^2}{16} = 1. \] Given that the point \((\alpha, 2\sqrt{5})\) lies on the hyperbola: \[ \frac{\alpha^2}{9} - \frac{(2\sqrt{5})^2}{16} = 1. \] Simplifying: \[ \frac{\alpha^2}{9} - \frac{20}{16} = 1 \Rightarrow \frac{\alpha^2}{9} = \frac{9}{4} \Rightarrow \alpha^2 = \frac{81}{4}. \] Thus, \[ \alpha = \pm \frac{9}{2}. \]

Step 5:
The focal distances from \((\alpha, 2\sqrt{5})\) to the foci \((\pm 5, 0)\) are: \[ d_1 = \sqrt{(\alpha + 5)^2 + (2\sqrt{5})^2}, \quad d_2 = \sqrt{(\alpha - 5)^2 + (2\sqrt{5})^2}. \] For \(\alpha = \frac{9}{2}\):
\[ d_1 = \sqrt{(9.5)^2 + 20} = \sqrt{110.25} = 10.5, \] \[ d_2 = \sqrt{(-0.5)^2 + 20} = \sqrt{20.25} = 4.5. \] Therefore, \[ p = d_1 \cdot d_2 = 10.5 \times 4.5 = 47.25. \]

Step 6:
\[ 4p = 4 \times 47.25 = 189. \]

Final Answer:
\[ \boxed{189} \]