Question

Ellipse \( E:\; \dfrac{x^2}{36} + \dfrac{y^2}{16} = 1 \). A hyperbola is confocal with the ellipse and the eccentricity of the hyperbola is equal to \(5\). If the principal axis of the hyperbola is the \(x\)-axis, then the length of the latus rectum of the hyperbola is:

• A. \( \dfrac{96}{\sqrt{5}} \)
• B. \( 24\sqrt{5} \)
• C. \( 18\sqrt{5} \)
• D. \( 12\sqrt{5} \)

Hint:

For confocal conics:
Confocal curves have the same focal distance \(c\)
Hyperbola eccentricity \( e = \frac{c}{a} \)
Latus rectum of hyperbola \( = \frac{2b^2}{a} \)

Answer 1

The Correct Option is A

Step 1: For the given ellipse: \[ a^2 = 36, b^2 = 16 \] \[ c^2 = a^2 - b^2 = 36 - 16 = 20 \Rightarrow c = \sqrt{20} \]
Step 2: Since the hyperbola is confocal with the ellipse, it has the same focal distance: \[ c = \sqrt{20} \] Given eccentricity of the hyperbola: \[ e = \frac{c}{a_h} = 5 \Rightarrow a_h = \frac{c}{5} = \frac{\sqrt{20}}{5} = \frac{2\sqrt{5}}{5} \]
Step 3: For the hyperbola: \[ b_h^2 = c^2 - a_h^2 = 20 - \frac{20}{25} = \frac{480}{25} \]
Step 4: Length of latus rectum of a hyperbola: \[ \text{LR} = \frac{2b_h^2}{a_h} \] \[ \text{LR} = \frac{2 \times \frac{480}{25}}{\frac{2\sqrt{5}}{5}} = \frac{960}{25} \cdot \frac{5}{2\sqrt{5}} = \frac{96}{\sqrt{5}} \]