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Step-by-Step Solution
Step 1: Identify the given elements of the first ellipse
The first ellipse is
$ \displaystyle E_{1} : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $
with $a > b$. The endpoints of its major axis are $ (\pm a, \, 0) $ and the endpoints of its minor axis are $ (0, \, \pm b) $. The foci of $E_{1}$ are at $ (\pm c_{1}, \, 0) $, where
$ \displaystyle c_{1}^2 = a^2 - b^2 \,.$
The eccentricity of $E_{1}$ is
$ \displaystyle e_{1} = \frac{c_{1}}{a} = \frac{\sqrt{a^2 - b^2}}{a}\,. $
Step 2: Determine the parameters of the second ellipse
Let the second ellipse be $ E_{2} $. We are told:
$E_{2}$ touches or passes through the endpoints of the major axis of $E_{1}$, i.e., $(\pm a, 0)$.
The foci of $E_{2}$ are the endpoints of the minor axis of $E_{1}$, namely $(0, \pm b)$.
Because the foci of $E_{2}$ lie along the $y$-axis, we can write its standard equation (centered at the origin) as
$ \displaystyle \frac{x^2}{B_{2}^2} + \frac{y^2}{A_{2}^2} = 1 \,,$
where $A_{2} > B_{2}$ and the foci of $E_{2}$ are $(0, \pm c_{2})$ with
$ \displaystyle c_{2}^2 = A_{2}^2 - B_{2}^2 \,.$
Step 3: Use the given conditions to find $A_{2}$ and $B_{2}$
Foci of $E_{2}$ are $(0, \pm b)$:
Since the foci are $(0, \pm b)$, we have $c_{2} = b$. Hence
$ \displaystyle b^2 = A_{2}^2 - B_{2}^2 \,.$
$E_{2}$ passes through $(\pm a, 0)$:
Plugging $x = \pm a$ and $y = 0$ into the equation of $E_{2}$ gives
$ \displaystyle \frac{a^2}{B_{2}^2} + \frac{0^2}{A_{2}^2} = 1 \;\;\Longrightarrow\;\; B_{2}^2 = a^2,\; \text{so} \; B_{2} = a. $
Thus,
$ \displaystyle B_{2} = a
\quad\text{and}\quad
c_{2}^2 = b^2 = A_{2}^2 - a^2 \;\;\Longrightarrow\;\; A_{2}^2 = a^2 + b^2. $
Step 4: Relate the eccentricities of $E_{1}$ and $E_{2}$
We are told that $E_{1}$ and $E_{2}$ have the same eccentricity. The eccentricity of $E_{2}$ is
$ \displaystyle e_{2} = \frac{c_{2}}{A_{2}} = \frac{b}{\sqrt{a^2 + b^2}}. $
Equating $e_{1}$ and $e_{2}$ gives
$ \displaystyle
\frac{\sqrt{a^2 - b^2}}{a} \;=\; \frac{b}{\sqrt{a^2 + b^2}}.
$
Step 5: Solve for the ratio $ \frac{a}{b} $
Square both sides to eliminate the square roots:
$ \displaystyle
\left(\frac{\sqrt{a^2 - b^2}}{a}\right)^2 \;=\; \left(\frac{b}{\sqrt{a^2 + b^2}}\right)^2.
$
$ \displaystyle
\frac{a^2 - b^2}{a^2} \;=\; \frac{b^2}{a^2 + b^2}.
$
$ \displaystyle
(a^2 - b^2)(a^2 + b^2) \;=\; a^2 b^2
\;\;\Longrightarrow\;\;
a^4 - b^4 \;=\; a^2 b^2.
$
Rewriting $ a^2 = X $ gives us the quadratic in $X$:
$ \displaystyle
X^2 - b^2 X - b^4 = 0.
$
Solving for $X$:
$ \displaystyle
X = \frac{b^2 \pm \sqrt{\,b^4 + 4b^4\,}}{2}
= \frac{b^2 \pm b^2\sqrt{5}}{2}
= \frac{b^2 (\,1 \pm \sqrt{5}\,)}{2}.
$
Since $a > b$, we choose the positive sign:
$ \displaystyle
a^2 = \frac{b^2 (\,1 + \sqrt{5}\,)}{2}.
$
Step 6: Find the common eccentricity
From above,
$ \displaystyle
e_{1}^2
= \frac{a^2 - b^2}{a^2}
= 1 - \frac{b^2}{a^2}
= 1 - \frac{b^2}{\frac{b^2 (1 + \sqrt{5})}{2}}
= 1 - \frac{2}{1 + \sqrt{5}}.
$
Rationalize $ \frac{2}{1 + \sqrt{5}} $:
$ \displaystyle
\frac{2}{1 + \sqrt{5}}
\times \frac{1 - \sqrt{5}}{1 - \sqrt{5}}
= \frac{2(\,1 - \sqrt{5}\,)}{(1 + \sqrt{5})(1 - \sqrt{5})}
= \frac{2(1 - \sqrt{5})}{1 - 5}
= \frac{2(1 - \sqrt{5})}{-4}
= \frac{\sqrt{5} - 1}{2}.
$
Hence,
$ \displaystyle
e_{1}^2 = 1 - \frac{\sqrt{5} - 1}{2}
= \frac{2 - (\sqrt{5} - 1)}{2}
= \frac{3 - \sqrt{5}}{2}.
$
Taking the positive square root (since eccentricity is positive),
$ \displaystyle
e_{1}
= \sqrt{\frac{3 - \sqrt{5}}{2}}
= \frac{\sqrt{5} - 1}{2},
$
because $ \left(\frac{\sqrt{5} - 1}{2}\right)^2 = \frac{3 - \sqrt{5}}{2}. $
Thus, the common eccentricity is
$ \displaystyle \frac{\sqrt{5} - 1}{2}. $
Final Answer
$ \displaystyle e = \frac{-1 + \sqrt{5}}{2}.
