Chapter 2, Problem 2.1P

To determine

The equation for the ellipse is given by $\left(\frac{x^2}{a^2}\right)+\left(\frac{y^2}{b^2}\right)=1$.

Answer to Problem 2.1P

The equation for the ellipse is $\left(\frac{x^2}{a^2}\right)+\left(\frac{y^2}{b^2}\right)=1$.

Explanation of Solution

The required diagram is shown in Figure 1.

Figure 1

Write the expression for $r$ by using Pythagoras theorem.

    $r^2={\left(x-ae\right)}^2+y^2$        (1)

Write the expression for $r^{\prime }$ by using Pythagoras theorem.

    ${r^{\prime }}^2={\left(x+ae\right)}^2+y^2$        (2)

Write the expression for the relation between $r$ and $r^{\prime }$.

    $r=2a-r^{\prime }$        (3)

Conclusion:

Substitute $\sqrt{{\left(x+ae\right)}^2+y^2}$ for $r^{\prime }$ equation (3).

    $\begin{array}{c}r=2a-\sqrt{{\left(x+ae\right)}^2+y^2}\\ r^2={\left(2a-\sqrt{{\left(x+ae\right)}^2+y^2}\right)}^2\end{array}$        (4)

Substitute ${\left(x-ae\right)}^2+y^2$ for $r^2$ equation (4).

    $\begin{array}{c}{\left(x-ae\right)}^2+y^2={\left(2a-\sqrt{{\left(x+ae\right)}^2+y^2}\right)}^2\\ x^2+a^2{e}^2-2aex+y^2=4a^2+{\left(x+ae\right)}^2-4a\sqrt{{\left(x+ae\right)}^2+y^2}+y^2\\ x^2+a^2{e}^2-2aex=4a^2+x^2+{(ae)}^2+2aex-4a\sqrt{{\left(x+ae\right)}^2+y^2}\\ 2a+2ex=2\sqrt{{\left(x+ae\right)}^2+y^2}\end{array}$

Solving further,

    $\begin{array}{c}a+ex=\sqrt{x^2+{\left(ae\right)}^2+2aex+y^2}\\ {\left(a+ex\right)}^2={\left(\sqrt{x^2+{\left(ae\right)}^2+2aex+y^2}\right)}^2\\ a^2+{\left(ex\right)}^2+2aex=x^2+{\left(ae\right)}^2+2aex+y^2\\ x^2\left(1-e^2\right)+y^2=a^2\left(1-e^2\right)\end{array}$

Solving further,

    $\frac{x^2}{a^2}+\frac{y^2}{a^2\left(1-e^2\right)}=1$        (5)

Substitute $a^2\left(1-e^2\right)$ for $b^2$ in equation (5).

    $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Thus, the equation for the ellipse is $\left(\frac{x^2}{a^2}\right)+\left(\frac{y^2}{b^2}\right)=1$.