# Ellipse 1 : E$_1 = \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 ( a > b)$ another ellipse 2 : $E_2$ which passes thru…

Problem :

Ellipse 1 : E$_1 = \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 ( a > b)$ another ellipse 2 : $E_2$ which passes thru extremities of major axis of $E_1$ and has its foci at ends of its minor axis. If eccentricity of both the ellipses are same, then find their eccentricity.

Since $E_2$ passes through extremities of major axis of $E_1$ therefore its length of minor axis is 2a; and coordinates of focus of $E_2$ = $2b$ but I am not getting further idea to solve this please help.

Let $e_1,e_2$ is the eccentricity of $E_1,E_2$ respectively.

First, we have $$e_1=\frac{\sqrt{a^2-b^2}}{a}.\tag1$$

Let $$E_2\ :\ \frac{x^2}{A^2}+\frac{y^2}{B^2}=1.$$ Here, note that $B\gt A.$

Since we have $$A=a,\ \ \ \sqrt{B^2-A^2}=b\Rightarrow B=\sqrt{a^2+b^2},$$ we have $$e_2=\frac{\sqrt{B^2-A^2}}{B}=\frac{b}{\sqrt{a^2+b^2}}.\tag2$$ Hence, from $(1),(2)$, we have $$\frac{\sqrt{a^2-b^2}}{a}=\frac{b}{\sqrt{a^2+b^2}}\Rightarrow a^2b^2=a^4-b^4$$$$\Rightarrow \left(\frac ba\right)^4+\left(\frac ba\right)^2-1=0\Rightarrow \left(\frac ba\right)^2=\frac{\sqrt 5-1}{2}.$$ Hence, the answer is $$\frac{\sqrt{a^2-b^2}}{a}=\sqrt{1-\left(\frac ba\right)^2}=\sqrt{1-\frac{\sqrt 5-1}{2}}=\sqrt{\frac{3-\sqrt 5}{2}}=\frac{\sqrt 5-1}{2}.$$

So, the equation of $E_2$ $$\frac{x^2}{p^2}+\frac{y^2}{a^2}=1$$ where $p$ is the major axis

So we have $b^2=a^2(1-e^2)\ \ \ \ (0)$ and $a^2=p^2(1-e^2)\ \ \ \ (1)$

As the coordinates of the foci are $(0\pm pe),b=pe$ where $e$ is the common eccentricity

So,from $(0), a^2(1-e^2)=(pe)^2\ \ \ \ (2)$

Divide $(1)$ by $(2)$