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 solve the problem, we need to find the eccentricity of an ellipse that passes through all four foci of the given ellipse and hyperbola. Let's start step-by-step.

1. First, determine the eccentricities of the given ellipse and hyperbola:
   • Ellipse: The given ellipse is \(\frac{x^2}{b^2} + \frac{y^2}{25} = 1\). The equation of an ellipse is generally of the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) where \(a \lt b\). Here, \(a^2 = b^2\) and \(b^2 = 25\).
   • The eccentricity \(e_1\) of the ellipse is given by \(e_1 = \sqrt{1 - \frac{a^2}{b^2}}\). Substituting, we get \(e_1 = \sqrt{1 - \frac{b^2}{25}}\).
   • Hyperbola: The given hyperbola is \(\frac{x^2}{16} - \frac{y^2}{b^2} = 1\). The standard form of a hyperbola is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).
   • The eccentricity \(e_2\) for the hyperbola is \(e_2 = \sqrt{1 + \frac{b^2}{a^2}}\), which gives \(e_2 = \sqrt{1 + \frac{b^2}{16}}\).
2. According to the problem, \(e_1 \cdot e_2 = 1\). Thus:
   • \(\sqrt{1 - \frac{b^2}{25}} \cdot \sqrt{1 + \frac{b^2}{16}} = 1\).
   • Squaring both sides, we get:
   • \(\left(1 - \frac{b^2}{25}\right) \cdot \left(1 + \frac{b^2}{16}\right) = 1\).
   • Expanding and simplifying, \(1 + \frac{b^2}{16} - \frac{b^2}{25} - \frac{b^4}{400} = 1\).
   • Solving, we get \(\frac{b^2}{16} - \frac{b^2}{25} = \frac{b^4}{400}\).
   • Simplifying, \(b^2 \left(\frac{1}{16} - \frac{1}{25}\right) = \frac{b^4}{400}\).
   • This leads to solving \(\frac{b^2 \cdot 9}{400} = \frac{b^4}{400}\) giving \(b^2 = 9\).
   • Thus, \(b = 3\) (since \(b \lt 5\)).
3. Now, substitute back 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, calculate the eccentricity of the new ellipse:
   • The new ellipse must pass through all four foci: \((\pm ae_1, 0)\) for the ellipse and \((\pm ae_2, 0)\) for the hyperbola.
   • The major axis is the line joining these foci, giving the total foci separation as \(2 \cdot 3 = 6\).
   • The eccentricity \(e = \frac{d}{2a} = \frac{3}{5}\).

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

Answer 2

To solve this problem, we need to determine the eccentricity \( e \) of an ellipse given certain conditions involving another ellipse and a hyperbola.

Step 1: Determining the Eccentricities of the Given Ellipse and Hyperbola

Consider the ellipse:

\(\frac{x^2}{b^2} + \frac{y^2}{25} = 1\)

Here, \( a = 5 \) and \( b < 5 \). The eccentricity of an ellipse is given by:

\(e_1 = \sqrt{1 - \frac{b^2}{a^2}}\)

For this ellipse, it becomes:

\(e_1 = \sqrt{1 - \frac{b^2}{25}}\)

Now consider the hyperbola:

\(\frac{x^2}{16} - \frac{y^2}{b^2} = 1\)

Here, \( a = 4 \). The eccentricity of a hyperbola is given by:

\(e_2 = \sqrt{1 + \frac{b^2}{a^2}}\)

For this hyperbola, it becomes:

\(e_2 = \sqrt{1 + \frac{b^2}{16}}\)

Given that \( e_1 e_2 = 1 \), we have:

\(\sqrt{1 - \frac{b^2}{25}} \times \sqrt{1 + \frac{b^2}{16}} = 1\)

Squaring both sides gives:

\(\left(1 - \frac{b^2}{25}\right)\left(1 + \frac{b^2}{16}\right) = 1\)

Expanding and solving for \( b^2 \), we get:

\(1 - \frac{b^2}{25} + \frac{b^2}{16} - \frac{b^4}{400} = 1\)

\(\frac{b^2}{16} - \frac{b^2}{25} = \frac{b^4}{400}\)

Simplifying, find \( b^2 \):

\(\frac{25b^2 - 16b^2}{400} = \frac{b^4}{400}\)

\(\frac{9b^2}{400} = \frac{b^4}{400}\)

Since \( b^2 \neq 0 \), we have:

\(b^2 = 9\)

Thus, \( b = 3 \), since \( b < 5 \).

Step 2: Eccentricity of the New Ellipse

The new ellipse passes through the four foci: two from the ellipse and two from the hyperbola.

For the ellipse \( \frac{x^2}{b^2} + \frac{y^2}{25} = 1 \) with \( b = 3 \), the foci are located at:

\((0, \pm 4)\)

For the hyperbola \( \frac{x^2}{16} - \frac{y^2}{b^2} = 1 \) with \( b^2 = 9 \), the foci are located at:

\((\pm 5, 0)\)

This means the foci are at points (0, 4), (0, -4), (5, 0), and (-5, 0).

The new ellipse, with axes along the coordinates and passing these points, has the equation:

\(\frac{x^2}{25} + \frac{y^2}{16} = 1\)

The eccentricity \( e \) of this ellipse is:

\(e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}}\)

Simplifying, we get:

\(e = \sqrt{\frac{9}{25}} = \frac{3}{5}\)

Thus, the eccentricity of the new ellipse is \(\frac{3}{5}\).