Question

The eccentricity of an ellipse \(E\) with centre at the origin \(O\) is \( \frac{\sqrt3}{2} \) and its directrices are \( x=\pm \frac{4\sqrt6}{3} \). Let \( H:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \) be a hyperbola whose eccentricity is equal to the length of semi-major axis of \(E\), and whose length of latus rectum is equal to the length of minor axis of \(E\). Then the distance between the foci of \(H\) is :

• A. \( \frac{4\sqrt2}{\sqrt7} \)
• B. \( \frac{4\sqrt2}{7} \)
• C. \( \frac{4}{\sqrt7} \)
• D. \( \frac{8}{7} \)

Answer 1

The Correct Option is A

Step 1: Understanding the Question
The problem involves two conic sections, an ellipse \(E\) and a hyperbola \(H\). We are given properties of the ellipse, which we must use to find its parameters. These parameters then define the properties of the hyperbola. Finally, we need to calculate the distance between the foci of the hyperbola.
Step 2: Key Formula or Approach
For an ellipse \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \):
- Directrix: \( x = \pm \frac{a}{e} \)
- Relation between parameters: \( b^2 = a^2(1-e^2) \)
- Length of minor axis: \( 2b \)
For a hyperbola \( \frac{x^2}{a_H^2}-\frac{y^2}{b_H^2}=1 \):
- Eccentricity relation: \( b_H^2 = a_H^2(e_H^2-1) \)
- Length of latus rectum: \( \frac{2b_H^2}{a_H} \)
- Foci are at \( (\pm c_H, 0) \) where \( c_H = a_H e_H \). Distance between foci is \( 2c_H \).
Step 3: Detailed Explanation
Part 1: Finding parameters of the Ellipse (E)
Given for ellipse E: eccentricity \( e_E = \frac{\sqrt{3}}{2} \) and directrix \( x = \frac{a_E}{e_E} = \frac{4\sqrt{6}}{3} \).
First, find the semi-major axis, \(a_E\).
\[ a_E = e_E \times \left( \frac{4\sqrt{6}}{3} \right) = \frac{\sqrt{3}}{2} \times \frac{4\sqrt{6}}{3} = \frac{4\sqrt{18}}{6} = \frac{2\sqrt{18}}{3} = \frac{2(3\sqrt{2})}{3} = 2\sqrt{2} \] Now, find the semi-minor axis, \(b_E\).
\[ b_E^2 = a_E^2 (1 - e_E^2) = (2\sqrt{2})^2 \left(1 - \left(\frac{\sqrt{3}}{2}\right)^2\right) = 8 \left(1 - \frac{3}{4}\right) = 8 \left(\frac{1}{4}\right) = 2 \] So, \( b_E = \sqrt{2} \).
Length of semi-major axis of E is \( a_E = 2\sqrt{2} \).
Length of minor axis of E is \( 2b_E = 2\sqrt{2} \).
Part 2: Finding parameters of the Hyperbola (H)
Let the hyperbola's parameters be \(a_H, b_H, e_H\).
Given: Eccentricity of H, \( e_H = \) length of semi-major axis of E.
\[ e_H = a_E = 2\sqrt{2} \] Given: Length of latus rectum of H = length of minor axis of E.
\[ \frac{2b_H^2}{a_H} = 2b_E = 2\sqrt{2} \implies \frac{b_H^2}{a_H} = \sqrt{2} \] We also have the relation for hyperbola eccentricity: \( b_H^2 = a_H^2(e_H^2 - 1) \).
\[ b_H^2 = a_H^2((2\sqrt{2})^2 - 1) = a_H^2(8 - 1) = 7a_H^2 \] Now we have two equations for \(b_H^2\): \( b_H^2 = a_H\sqrt{2} \) and \( b_H^2 = 7a_H^2 \).
Equating them: \[ a_H\sqrt{2} = 7a_H^2 \] Since \(a_H \neq 0\), we can divide by \(a_H\): \[ \sqrt{2} = 7a_H \implies a_H = \frac{\sqrt{2}}{7} \] Now we find \(c_H\), the distance from the center to a focus of the hyperbola.
\[ c_H = a_H e_H = \left(\frac{\sqrt{2}}{7}\right) (2\sqrt{2}) = \frac{4}{7} \] The distance between the foci of the hyperbola is \( 2c_H \).
\[ \text{Distance} = 2 \times \frac{4}{7} = \frac{8}{7} \] Step 4: Final Answer
The calculated distance is \( \frac{8}{7} \), which is option (D).