conic sections
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.
We may suppose that $$\text{the ellipse$\ :\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\quad a\gt b\gt 0$}$$ $$\text{the circle$\ :\ x^2+y^2=a^2-b^2$}$$
As you wrote, we have $$2a=17\quad\Rightarrow \quad a=\frac{17}{2}$$
Since $$\frac{a^2-b^2-y^2}{a^2}+\frac{y^2}{b^2}=1\quad\Rightarrow\quad |y|=\frac{b^2}{\sqrt{a^2-b^2}}$$
we have $$30=\frac 12\times 2\sqrt{a^2-b^2}\times \frac{b^2}{\sqrt{a^2-b^2}}\quad\Rightarrow\quad b=\sqrt{30}.$$
Thus, the answer is $$2\sqrt{a^2-b^2}=2\sqrt{\left(\frac{17}{2}\right)^2-30}=\color{red}{13}.$$
Hint:
Without loss of generality, one of the ellipses is the unit circle centered at the origin. (If not, you can translate its center to the origin, counter-rotate to bring its major axis horizontal and rescale non-isotropically by the axis lengths).
Then the problem reduces to checking if an ellipse is wholly contained in the unit circle.
Let the parametric equation of that ellipse be
$$\vec p=\vec p_c+\vec a\cos t+\vec b\sin t.$$ We need to guarantee the inequality
$$\vec p^2\le 1,$$ i.e.
$$\vec p_c^2+\vec a^2\cos^2t+\vec b^2\sin^2t+2\vec p_c\vec a\cos t+2\vec p_c\vec b\sin t\le 1$$ (we have $\vec a\vec b=0$).
The inequality is ensured by finding the maxima of the trigonometric polynomial and showing the they don't exceed $1$.
Unfortunately, this leads to a quartic equation, so that an analytical solution promises to be painful. The problem is very close to that of finding the shortest distance of a point to an ellipse, or to ellipse offsetting.
There is an analytical solution, but it is a little scary.
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$$
"deflate" the circle while you "inflate" the ellipse; that means that you reduce the radius of the circle until it reduces to a point, while you compute the corresponding offset curve of the ellipse (https://en.wikipedia.org/wiki/Parallel_curve).
the offset curve has a known implicit equation, given in "Brief Atlas of Offset Curves, Juana Sandra", available online.
Best Answer
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 $$\frac{\sqrt{a^2-b^2}}{a}=\sqrt{1-\left(\frac ba\right)^2}=\sqrt{1-\frac{\sqrt 5-1}{2}}=\sqrt{\frac{3-\sqrt 5}{2}}=\frac{\sqrt 5-1}{2}.$$