Chapter 8, Problem 4TP

Solution

To calculate: The equation of the ellipse

The equation of the ellipse is $\frac{{(x-9/2)}^2}{121}+\frac{y^2}{403/4}=1$

Given information:

Measures of the radii are the consecutive integers from $1to9$

Calculation:

If the intersection points lie on an ellipse, then the major axis of that ellipse must be

  $2a=1+9+1=11$

Because that's the distance between the two furthest points $(-1,0)$ and $(10,0)$ . The circles are centered at $C_1=(1,0)$ and $C_2=(8,0)$ .

Let $T=\left(x_T,y_T\right)$ be a point which is the intersection of the circles centered at $C_1$ with radius $1\le m\le 9$ and at $C_2$ with radius $1\le n\le 9$ . Then

  $\begin{array}{c}{\left(x_T-1\right)}^2+y_T^2=m^2\\ {\left(x_T-8\right)}^2+y_T^2=n^2\end{array}$

Then

  $‖T{C}_1‖+‖T{C}_2‖=m+n$

Since only consider the intersection of the circle of radius $m$ with the circle of radius $11-m$ ,

  $‖T{C}_1‖+‖T{C}_2‖=m+11-m=11$

Hence, the sum of the distances of $T$ from $C_1$ and $C_2$ is the same for all intersections $T$ , so they all lie on the ellipse with foci $C_1$ and $C_2$ .

The center $C=\left(x_0,y_0\right)$ of the ellipse is the midpoint of $\overline{C_1{C}_2}$ , so

  $\begin{array}{c}{x}_0=\frac{1+8}{2}=\frac{9}{2},\\ y_0=0.\end{array}$

The linear eccentricity is $\begin{array}{c}c=‖C_{1}C‖\\ =\frac{9}{2}\end{array}$

and from

  $c^2=a^2-b^2$

Then

  $\begin{array}{c}b=\sqrt{a^2-c^2}\\ =\sqrt{{11}^2-{(9/2)}^2}\\ =\frac{\sqrt{121\cdot 4-81}}{2}\\ =\frac{\sqrt{403}}{2}\end{array}$

Now the equation of the ellipse is

  $\frac{{(x-9/2)}^2}{121}+\frac{y^2}{403/4}=1$