Question

Let $ e_1 $ and $ e_2 $ be the eccentricities of the ellipse $ \frac{x^2}{b^2} + \frac{y^2}{25} = 1 $ and the hyperbola $ \frac{x^2}{16} - \frac{y^2}{b^2} = 1, $ respectively. If $ b<5 $ and $ e_1 e_2 = 1 $, then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is:

• A. \( \frac{4}{5} \)
• B. \( \frac{3}{5} \)
• C. \( \frac{\sqrt{7}}{4} \)
• D. \( \frac{\sqrt{3}}{2} \)

Hint:

When given the product of eccentricities of two conic sections, you can compute the eccentricity of a new ellipse by taking the geometric mean of the given eccentricities.

Answer 1

The Correct Option is B

We are given the equations of an ellipse and a hyperbola: - For the ellipse: \[ \frac{x^2}{b^2} + \frac{y^2}{25} = 1, \] where \( a^2 = 25 \) and \( b^2 \) is the semi-minor axis squared. The eccentricity \( e_1 \) of the ellipse is: \[ e_1 = \sqrt{1 - \frac{b^2}{25}}. \] - For the hyperbola: \[ \frac{x^2}{16} - \frac{y^2}{b^2} = 1, \] where \( a^2 = 16 \) and \( b^2 \) is the semi-minor axis squared. The eccentricity \( e_2 \) of the hyperbola is: \[ e_2 = \sqrt{1 + \frac{b^2}{16}}. \] 
Step 1: Use the Given Condition \( e_1 e_2 = 1 \)
We are told that \( e_1 e_2 = 1 \). So, we can write: \[ \sqrt{1 - \frac{b^2}{25}} \times \sqrt{1 + \frac{b^2}{16}} = 1. \] Squaring both sides: \[ \left( 1 - \frac{b^2}{25} \right) \times \left( 1 + \frac{b^2}{16} \right) = 1. \] Expanding the left side: \[ 1 - \frac{b^2}{25} + \frac{b^2}{16} - \frac{b^4}{400} = 1. \] Simplifying: \[ - \frac{b^2}{25} + \frac{b^2}{16} = \frac{b^4}{400}. \] Taking the common denominator for \( b^2 \) terms: \[ \frac{-16b^2 + 25b^2}{400} = \frac{b^4}{400}. \] Simplifying: \[ \frac{9b^2}{400} = \frac{b^4}{400}. \] Multiplying both sides by 400: \[ 9b^2 = b^4. \] Solving for \( b^2 \): \[ b^4 - 9b^2 = 0 \quad \Rightarrow \quad b^2(b^2 - 9) = 0. \] Thus, \( b^2 = 9 \) (since \( b \neq 0 \)). 
Step 2: Calculate the Eccentricity of the Ellipse
Now substitute \( b^2 = 9 \) into the equation for \( e_1 \): \[ e_1 = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}. \] 
Step 3: Calculate the Eccentricity of the Hyperbola
Now substitute \( b^2 = 9 \) into the equation for \( e_2 \): \[ e_2 = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}. \] 
Step 4: Find the Eccentricity of the Required Ellipse
We are now asked to find the eccentricity of the ellipse that passes through all four foci (two of the ellipse and two of the hyperbola). 
This ellipse has its axes along the coordinate axes. 
The product of the eccentricities of the ellipse and hyperbola is given as \( e_1 e_2 = 1 \). 
The eccentricity of the new ellipse will be the geometric mean of the two eccentricities: \[ e = \sqrt{e_1 e_2} = \sqrt{\frac{4}{5} \times \frac{5}{4}} = \frac{3}{5}. \] Thus, the eccentricity of the required ellipse is \( \frac{3}{5} \), which corresponds to option (2).