Question

Ellipse \( E: \frac{x^2}{36} + \frac{y^2}{25} = 1 \), A hyperbola confocal with ellipse \( E \) and eccentricity of hyperbola is equal to 5. The length of latus rectum of hyperbola is, if principle axis of hyperbola is x-axis?

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

Hint:

For hyperbolas, use the relationship between eccentricity and the axes to find latus rectum length. For confocal ellipses and hyperbolas, eccentricity plays a key role in determining dimensions.

Answer 1

The Correct Option is A

To find the length of the latus rectum of the hyperbola confocal with the given ellipse \(E: \frac{x^2}{36} + \frac{y^2}{25} = 1\), we need to first determine the parameters of the hyperbola and then calculate the latus rectum.

1. \(E\) is an ellipse with the equation \(\frac{x^2}{36} + \frac{y^2}{25} = 1\). The semi-major axis \(a\) is 6, and the semi-minor axis \(b\) is 5.
2. Since the hyperbola is confocal with the ellipse, both share the same foci. Thus, the distance of the focus from the center for the ellipse is given by \(c = \sqrt{a^2 - b^2} = \sqrt{36 - 25} = \sqrt{11}\).
3. The hyperbola is given an eccentricity \(e = 5\). The relationship between the eccentricity, semi-major axis (\(a_h\)), and the distance of focus (\(c_h\)) for a hyperbola is \(e = \frac{c_h}{a_h}\).
4. Since the hyperbola is confocal with the ellipse, \(c_h = c = \sqrt{11}\). Therefore, the equation becomes: \(5 = \frac{\sqrt{11}}{a_h} \Rightarrow a_h = \frac{\sqrt{11}}{5}\).
5. The latus rectum of the hyperbola is given by the formula \(\frac{2b_h^2}{a_h}\). First, calculate \(b_h\) using: \(b_h^2 = c_h^2 - a_h^2 = 11 - \left(\frac{\sqrt{11}}{5}\right)^2\).
6. Since \(b_h^2 = c_h^2 - a_h^2 \Rightarrow b_h^2 = 11 - \frac{11}{25} = \frac{11 \times 24}{25} = \frac{264}{25}\), hence \(b_h^2 = \frac{264}{25}\).
7. Now, calculate the latus rectum, which is \(\frac{2b_h^2}{a_h} = \frac{2 \times \frac{264}{25}}{\frac{\sqrt{11}}{5}} = \frac{528}{\sqrt{11}}\).
8. Simplify this to get the length of the latus rectum as \(\frac{528 \times 5}{\sqrt{11} \times 5} = \frac{96}{\sqrt{5}}\).

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