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

The foci of the ellipse are given by the equation \(\frac{x^2}{36} + \frac{y^2}{16} = 1\). Here, the semi-major axis \(a = \sqrt{36} = 6\), and the semi-minor axis \(b = \sqrt{16} = 4\).

The eccentricity \(e\) for this ellipse is calculated using the formula \(e = \sqrt{1 - \frac{b^2}{a^2}}\). 

Substituting the given values, we have:

\[e = \sqrt{1 - \frac{4^2}{6^2}} = \sqrt{1 - \frac{16}{36}} = \sqrt{\frac{20}{36}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}\]

The foci of the ellipse are located at \((\pm ae, 0) = \left(\pm 6 \times \frac{\sqrt{5}}{3}, 0\right) = (\pm 2\sqrt{5}, 0)\).

The foci of the hyperbola also coincide at these points. The eccentricity of the hyperbola is given as \(e = 5\).

For a hyperbola, the relation between the semi-major axis \(a\) and the distance of the foci \(\pm c\) is given by \(c = ae\).

The distance of each focus from the center is also given by \(c = 2\sqrt{5}\).

So, we have the equation:

\[2\sqrt{5} = 5a\]

Solving for \(a\),

\[a = \frac{2\sqrt{5}}{5}\]

The length of the latus rectum for a hyperbola is given by the formula \(\frac{2b^2}{a}\). We need to find \(b\) using the relationship \(c^2 = a^2 + b^2\).

Given \(c = 2\sqrt{5}\):

\[(2\sqrt{5})^2 = a^2 + b^2 \implies 20 = \left(\frac{2\sqrt{5}}{5}\right)^2 + b^2\]\[20 = \frac{20}{25} + b^2 \implies 20 = \frac{4}{5} + b^2\]\[b^2 = 20 - \frac{4}{5} = \frac{100}{5} - \frac{4}{5} = \frac{96}{5}\]

The length of the latus rectum is:

\[\frac{2b^2}{a} = \frac{2 \times \frac{96}{5}}{\frac{2\sqrt{5}}{5}} = \frac{192}{5} \times \frac{5}{2\sqrt{5}} = \frac{96}{\sqrt{5}}\]

Hence, the length of the latus rectum is \(\frac{96}{\sqrt{5}}\).