Let E_1:x^2a^2+y^2b^2=1, ab.

Question

Let $E_1:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, $a>b$. Let $E_2$ be another ellipse such that it touches the endpoints of the major axis of $E_1$ and the foci of $E_2$ are the endpoints of the minor axis of $E_1$. If $E_1$ and $E_2$ have the same eccentricity, then its value is

Accepted Answer

$E_2$ has major axis along $y$ with foci at $(0,\pm b)$ (so $c_2=b$) and semi-minor axis $a$ (passing through $(\pm a,0)$), so its semi-major axis $B_2$ satisfies $B_2^2=a^2+b^2$.

$$e_1^2=1-\frac{b^2}{a^2},\qquad e_2^2=\frac{c_2^2}{B_2^2}=\frac{b^2}{a^2+b^2}.$$

Let $r=\dfrac{b^2}{a^2}$; then $e_1^2=1-r$ and $e_2^2=\dfrac r{1+r}$. Setting $e_1=e_2$:
$$1-r=\frac r{1+r}\ \Longrightarrow\ (1-r)(1+r)=r\ \Longrightarrow\ r^2+r-1=0\ \Longrightarrow\ r=\frac{-1+\sqrt5}2.$$

Then $e^2=1-r=\dfrac{3-\sqrt5}2$, which satisfies $e^2+e-1=0$, giving

$$e=\frac{-1+\sqrt5}2.$$

Solutions · 1