Question

If the product of the eccentricities of the ellipse \( \frac{x^2}{16} + \frac{y^2}{b^2} = 1 \) and the hyperbola \( \frac{x^2}{9} - \frac{y^2}{16} = 1 \) is 1, then the value of \( b^2 \) is:

• A. \( \frac{12}{25} \)
• B. \( 144 \)
• C. \( 25 \)
• D. \( \frac{144}{25} \)

Hint:

For an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), the eccentricity is: \( e = \sqrt{1 - \frac{b^2}{a^2}}. \) For a hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the eccentricity is: \( e = \sqrt{1 + \frac{b^2}{a^2}}. \)

Answer 1

The Correct Option is D

Step 1: Find the eccentricity of the ellipse For the ellipse: \( \frac{x^2}{16} + \frac{y^2}{b^2} = 1. \) The eccentricity of an ellipse is given by: \( e = \sqrt{1 - \frac{b^2}{a^2}}. \) Here, \( a^2 = 16 \), so: \( e_1 = \sqrt{1 - \frac{b^2}{16}}. \) Step 2: Find the eccentricity of the hyperbola For the hyperbola: \( \frac{x^2}{9} - \frac{y^2}{16} = 1. \) The eccentricity of a hyperbola is given by: \( e = \sqrt{1 + \frac{b^2}{a^2}}. \) Here, \( a^2 = 9 \), and \( b^2 = 16 \), so: \( e_2 = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}. \) Step 3: Using the given condition We are given that: \( e_1 \times e_2 = 1. \) Substituting the values: \( \sqrt{1 - \frac{b^2}{16}} \times \frac{5}{3} = 1. \) Squaring both sides: \( \left(1 - \frac{b^2}{16}\right) \times \frac{25}{9} = 1. \) \( \frac{25}{9} - \frac{25b^2}{144} = 1. \) Multiplying by 144 to clear fractions: \( 400 - 25b^2 = 144. \) \( 25b^2 = 256. \) \( b^2 = \frac{144}{25}. \)