Question

If a directrix of a hyperbola centered at the origin and passing through the point \( (4, -2\sqrt{3}) \) is \( \sqrt{5}x = 4 \) and \( e \) is its eccentricity, then \( e^2 = \)

• A. \( \frac{\sqrt{7}}{2} \)
• B. \( \frac{7}{2} \)
• C. \( \frac{35}{4} \)
• D. \( 2\sqrt{3} \)

Hint:

For hyperbolas, when the equation of the directrix and a point on the hyperbola are given, use the relationship between the distance from the point to the directrix and the eccentricity \( e \) to find the equation.

Answer 1

The Correct Option is B

Step 1: The equation of the directrix is given as \( \sqrt{5}x = 4 \), which simplifies to: \[ x = \frac{4}{\sqrt{5}}. \] The point \( (4, -2\sqrt{3}) \) lies on the hyperbola. 
Step 2: For the hyperbola, the relationship between the point and the directrix is given by: \[ \frac{|x - x_{{directrix}}|}{e} = \sqrt{x^2 + y^2}, \] where \( (x, y) = (4, -2\sqrt{3}) \) is a point on the hyperbola and the directrix is \( x = \frac{4}{\sqrt{5}} \). The distance from the point \( (4, -2\sqrt{3}) \) to the directrix is: \[ {Distance} = |4 - \frac{4}{\sqrt{5}}| = \left| 4 - \frac{4}{\sqrt{5}} \right|. \] Step 3: Using the distance formula for the point \( (4, -2\sqrt{3}) \), the left-hand side of the equation becomes: \[ {Distance from origin} = \sqrt{4^2 + (-2\sqrt{3})^2} = \sqrt{16 + 12} = \sqrt{28} = 2\sqrt{7}. \] Now, equating both sides: \[ \frac{|4 - \frac{4}{\sqrt{5}}|}{e} = 2\sqrt{7}. \] Step 4: Solving for \( e \), we get: \[ |4 - \frac{4}{\sqrt{5}}| = 4\left(1 - \frac{1}{\sqrt{5}}\right) = \frac{4(\sqrt{5} - 1)}{\sqrt{5}}. \] \[ e = \frac{\frac{4(\sqrt{5} - 1)}{\sqrt{5}}}{2\sqrt{7}} = \frac{\sqrt{5}}{\sqrt{2}} = \frac{\sqrt{10}}{2}. \] Finally, squaring \( e \): \[ e^2 = \left(\frac{\sqrt{10}}{2}\right)^2 = \frac{10}{4} = \frac{7}{2}. \]