Let a denote the semi-major axis, b the semi-minor axis, and 2c the distance between the foci of the ellipse. The eccentricity e is defined as \( e = \frac{c}{a} \).
Given that the length of the minor axis is half the distance between the foci, we have:
\[ 2b = \frac{1}{2} \times 2c \Rightarrow 2b = c \]
Substituting \( c = ae \) into the equation yields:
\[ 2b = ae \]
Utilizing the relationship \( b = a\sqrt{1 - e^2} \), we substitute for b:
\[ 2a\sqrt{1 - e^2} = ae \]
Dividing by a gives:
\[ 2\sqrt{1 - e^2} = e \]
Squaring both sides results in:
\[ 4(1 - e^2) = e^2 \]
Expanding and rearranging the terms leads to:
\[ 4 = 5e^2 \]
\[ e^2 = \frac{4}{5} \]
\[ e = \frac{2}{\sqrt{5}} \]
Let each of the two ellipses $E_1:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\;(a>b)$ and $E_2:\dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}=1A$