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

To address the problem, we must determine the eccentricity of an ellipse that encompasses all four foci of the provided ellipse and hyperbola. We will proceed sequentially.

1. Initially, ascertain the eccentricities of the given ellipse and hyperbola:
   • Ellipse: The provided ellipse is \( \frac{x^2}{b^2} + \frac{y^2}{25} = 1 \). The standard ellipse equation is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( a \lt b \). In this case, \( a^2 = b^2 \) and \( b^2 = 25 \).
   • The ellipse's eccentricity, denoted as \( e_1 \), is calculated using \( e_1 = \sqrt{1 - \frac{a^2}{b^2}} \). Substituting the given values yields \( e_1 = \sqrt{1 - \frac{b^2}{25}} \).
   • Hyperbola: The given hyperbola is \( \frac{x^2}{16} - \frac{y^2}{b^2} = 1 \). The standard hyperbola form is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
   • The hyperbola's eccentricity, \( e_2 \), is given by \( e_2 = \sqrt{1 + \frac{b^2}{a^2}} \), resulting in \( e_2 = \sqrt{1 + \frac{b^2}{16}} \).
2. As per the problem statement, \( e_1 \cdot e_2 = 1 \). This leads to:
   • \( \sqrt{1 - \frac{b^2}{25}} \cdot \sqrt{1 + \frac{b^2}{16}} = 1 \).
   • Squaring both sides gives:
   • \( \left(1 - \frac{b^2}{25}\right) \cdot \left(1 + \frac{b^2}{16}\right) = 1 \).
   • Expanding and simplifying results in \( 1 + \frac{b^2}{16} - \frac{b^2}{25} - \frac{b^4}{400} = 1 \).
   • Rearranging to solve for \( b^2 \): \( \frac{b^2}{16} - \frac{b^2}{25} = \frac{b^4}{400} \).
   • Further simplification yields \( b^2 \left(\frac{1}{16} - \frac{1}{25}\right) = \frac{b^4}{400} \).
   • This simplifies to \( \frac{b^2 \cdot 9}{400} = \frac{b^4}{400} \), from which \( b^2 = 9 \).
   • Therefore, \( b = 3 \) (since \( b \lt 5 \)).
3. Now, substitute the value of \( b \) to find the eccentricities:
   • \( e_1 = \sqrt{1 - \frac{3^2}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \).
   • \( e_2 = \sqrt{1 + \frac{3^2}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4} \).
4. Finally, compute the eccentricity of the new ellipse:
   • The new ellipse must pass through all four foci, which are \( (\pm ae_1, 0) \) for the ellipse and \( (\pm ae_2, 0) \) for the hyperbola.
   • The major axis connects these foci, resulting in a total focal separation of \( 2 \cdot 3 = 6 \).
   • The eccentricity is calculated as \( e = \frac{d}{2a} = \frac{3}{5} \).

Consequently, the eccentricity of the required ellipse is \( \frac{3}{5} \).