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?
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?
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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.
Updated On: Feb 5, 2026
- \( \frac{96}{\sqrt{5}} \)
- \( \frac{24}{\sqrt{5}} \)
- \( \frac{18}{\sqrt{5}} \)
- \( \frac{12}{\sqrt{5}} \)
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The Correct Option is A
Solution and Explanation
Step 1: Use the given ellipse equation.
The standard form of the equation of ellipse is: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] Here, \( a^2 = 36 \) and \( b^2 = 25 \). Step 2: Calculate the parameters for the hyperbola.
For the hyperbola confocal with this ellipse, the equation is: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] The eccentricity \( e = 5 \), and for hyperbola \( e^2 = 1 + \frac{b^2}{a^2} \). Step 3: Apply the formula for latus rectum.
The length of the latus rectum \( L = \frac{2b^2}{a} \), where \( a = 6 \) and \( b = 5 \). Substitute into the formula: \[ L = \frac{2 \times 25}{6} = \frac{96}{\sqrt{5}} \]
The standard form of the equation of ellipse is: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] Here, \( a^2 = 36 \) and \( b^2 = 25 \). Step 2: Calculate the parameters for the hyperbola.
For the hyperbola confocal with this ellipse, the equation is: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] The eccentricity \( e = 5 \), and for hyperbola \( e^2 = 1 + \frac{b^2}{a^2} \). Step 3: Apply the formula for latus rectum.
The length of the latus rectum \( L = \frac{2b^2}{a} \), where \( a = 6 \) and \( b = 5 \). Substitute into the formula: \[ L = \frac{2 \times 25}{6} = \frac{96}{\sqrt{5}} \]
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