Let E_1: x^29+y^24=1.

Question

Let $E_1: \dfrac{x^2}{9}+\dfrac{y^2}{4}=1$. Ellipses $E_i$ are constructed such that their centres and eccentricities are the same as that of $E_1$, and the length of the minor axis of $E_i$ equals the length of the major axis of $E_{i+1}$. If $A_i$ is the area of the ellipse $E_i$, then $\dfrac{5}{\pi}\displaystyle\sum_{i=1}^{\infty}A_i$ is equal to \_\_\_\_\_\_\_\_.

Answer: 54

Accepted Answer

For $E_1$: $a_1=3,\ b_1=2$, so $e=\sqrt{1-\tfrac49}=\dfrac{\sqrt5}{3}$, constant for all $E_i$.

Given $2b_i=2a_{i+1}$, i.e. $a_{i+1}=b_i$, and since eccentricity is fixed, $b_{i+1}=b_i\sqrt{1-e^2}=b_i\cdot\dfrac23$.

$$A_i=\pi a_i b_i,\qquad \frac{A_{i+1}}{A_i}=\frac{a_{i+1}b_{i+1}}{a_ib_i}=\frac{b_i\cdot\frac23b_i}{a_ib_i}=\frac{2b_i}{3a_i}.$$

Since $b_1/a_1=2/3$ and the ratio $b_i/a_i$ stays $2/3$ at every stage (as both scale by the same factor), the area ratio is the constant $r=\dfrac{2}{3}\cdot\dfrac23=\dfrac49$.

$$A_1=\pi(3)(2)=6\pi,\qquad \sum_{i=1}^\infty A_i=\frac{6\pi}{1-4/9}=\frac{54\pi}{5}.$$

$$\frac{5}{\pi}\sum A_i = \frac{5}{\pi}\cdot\frac{54\pi}{5}=54.$$

Solutions · 1