EllipseHardBloom L3

Question

A series of ellipses $E_{1},E_{2},\ldots\ldots E_{n}$ are drawn such that $E_{n}$ touches $E_{n - 1}$ at the extremeties of the major axis of $E_{n - 1}$ and the foci of $E_{n}$ coincide with the extremeties of minor axis of $E_{n - 1}$. If ecentricity of the ellipses is independent of $n$, then the value of ecentricity is :

Options

A.$\frac{\sqrt{5}}{3}$
B.$\frac{\sqrt{5} + 1}{4}$
C.$\frac{\sqrt{5} - 1}{2}$
D.$\frac{\sqrt{5} - 1}{3}$

Solution

Sol. $\begin{matrix} & \ \\ & \\ & \\ & \\ & \\ b^{'} & \ = a,a^{'}e = b \\ b^{'2} & \ = a^{'2} - a^{'2}e^{2} \\ a^{2} & \ = \frac{b^{2}}{e^{2}} - b^{2} \end{matrix}$

$$\begin{matrix} & e^{2} = \frac{b^{2}}{a^{2}}\left( 1 - e^{2} \right) = \left( 1 - e^{2} \right)^{2} \\ & 1 - e^{2} = \pm e \\ & e^{2} + e - 1 = 0 \\ \text{~}\text{or}\text{~} & e^{2} - e - 1 = 0 \\ \Rightarrow & e = \frac{- 1 \pm \sqrt{5}}{2} \\ \text{~}\text{or}\text{~} & \frac{1 \pm \sqrt{5}}{2} \\ \Rightarrow & e = \frac{\sqrt{5} - 1}{2} \end{matrix}$$

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