Question

Let the foci of a hyperbola coincide with the foci of the ellipse \(\frac{x^2}{36} + \frac{y^2}{16} = 1\). If the eccentricity of the hyperbola is 5, then the length of its latus rectum is:

• A. \(24\sqrt{5}\)
• B. 12
• C. 16
• D. \(\frac{96}{\sqrt{5}}\)

Hint:

Confocal conics problems are common. The key is to remember that they share the same 'c' value (distance from center to focus).
For an ellipse, \(c^2 = a^2 - b^2\).
For a hyperbola, \(c^2 = A^2 + B^2 = (Ae_H)^2\).
Equating the 'c' values is the first step that connects the two conics.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given an ellipse and a hyperbola that are confocal (share the same foci). We are also given the eccentricity of the hyperbola. We need to find the length of the latus rectum of this hyperbola.
Step 2: Key Formula or Approach:
1. Ellipse Properties: For an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) (\(a>b\)), the distance of the foci from the center is \(c_E\), where \(c_E^2 = a^2 - b^2\). The foci are at \((\pm c_E, 0)\).
2. Hyperbola Properties: For a hyperbola \(\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1\), the distance of the foci from the center is \(c_H\), where \(c_H = Ae_H\) and \(c_H^2 = A^2 + B^2\). The foci are at \((\pm c_H, 0)\). \(e_H\) is the eccentricity.
3. Latus Rectum of Hyperbola: The length of the latus rectum is given by the formula \(\frac{2B^2}{A}\).
Step 3: Detailed Explanation:
Part 1: Find the foci of the ellipse.
The ellipse is \(\frac{x^2}{36} + \frac{y^2}{16} = 1\). Here, \(a^2 = 36\) and \(b^2 = 16\). The distance of the foci from the center is \(c_E\), where: \[ c_E^2 = a^2 - b^2 = 36 - 16 = 20 \] \[ c_E = \sqrt{20} = 2\sqrt{5} \] So, the foci of the ellipse are at \((\pm 2\sqrt{5}, 0)\). Part 2: Use confocal property to find hyperbola parameters.
Since the hyperbola is confocal with the ellipse, its foci are also at \((\pm 2\sqrt{5}, 0)\). For the hyperbola, the distance of the foci from the center is \(c_H = 2\sqrt{5}\). Let the hyperbola's equation be \(\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1\). We know that for a hyperbola, \(c_H = Ae_H\), where \(e_H\) is its eccentricity. We are given \(e_H = 5\). \[ 2\sqrt{5} = A \cdot 5 \] \[ A = \frac{2\sqrt{5}}{5} \] Now we find \(B^2\). The relationship for a hyperbola is \(B^2 = A^2(e_H^2 - 1)\). \[ A^2 = \left(\frac{2\sqrt{5}}{5}\right)^2 = \frac{4 \times 5}{25} = \frac{20}{25} = \frac{4}{5} \] \[ B^2 = \frac{4}{5} (5^2 - 1) = \frac{4}{5} (25 - 1) = \frac{4}{5} \times 24 = \frac{96}{5} \] Part 3: Calculate the length of the latus rectum.
The length of the latus rectum of the hyperbola is \(\frac{2B^2}{A}\). \[ L.R. = \frac{2 \times (96/5)}{2\sqrt{5}/5} = \frac{192/5}{2\sqrt{5}/5} = \frac{192}{2\sqrt{5}} = \frac{96}{\sqrt{5}} \] Step 4: Final Answer:
The length of the latus rectum of the hyperbola is \(\frac{96}{\sqrt{5}}\).