Chapter 8.2, Problem 51E

Solution

To Prove: An equation for the ellipse with the given center, the given foci, and the given semi-major axis is $\frac{y^2}{a^2}+\frac{x^2}{b^2}=1$ , where $b^2=a^2-c^2$ .

It has been shown that an equation for the ellipse with the given center, the given foci, and the given semi-major axis is $\frac{y^2}{a^2}+\frac{x^2}{b^2}=1$ , where $b^2=a^2-c^2$ .

Given:

Center $\left(0,0\right)$ , foci $\left(0,\pm c\right)$ , and semi-major axis $a>c\ge 0$ .

Concept used:

The sum of the distances from the foci to each point on the ellipse is a constant.

Calculation:

It is given that an ellipse has center $\left(0,0\right)$ , foci $\left(0,\pm c\right)$ , and semi-major axis $a>c\ge 0$ .

Then, $F_1\left(0,c\right)$ and $F_2\left(0,-c\right)$ are the foci.

Then, the ellipse is defined as the set of points $P\left(x,y\right)$ such that $P{F}_1+P{F}_2=2a$ .

Applying the distance formula,

  $P{F}_1=\sqrt{{\left(x-0\right)}^2+{\left(y-c\right)}^2}$ ,

  $P{F}_2=\sqrt{{\left(x-0\right)}^2+{\left(y-\left(-c\right)\right)}^2}$

Put these expressions in $P{F}_1+P{F}_2=2a$ to get,

  $\begin{array}{c}\sqrt{{\left(x-0\right)}^2+{\left(y-c\right)}^2}+\sqrt{{\left(x-0\right)}^2+{\left(y-\left(-c\right)\right)}^2}=2a\\ \sqrt{x^2+{\left(y-c\right)}^2}+\sqrt{x^2+{\left(y+c\right)}^2}=2a\\ \sqrt{x^2+{\left(y-c\right)}^2}=2a-\sqrt{x^2+{\left(y+c\right)}^2}\end{array}$

Squaring both sides,

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

Squaring both sides,

  $\begin{array}{c}{x}^2+{\left(y+c\right)}^2={\left(\frac{cy}{a}+a\right)}^2\\ x^2+y^2+c^2+2cy=\frac{c^2{y}^2}{a^2}+a^2+2cy\\ x^2+y^2+c^2=\frac{c^2{y}^2}{a^2}+a^2\end{array}$

It is given that $b^2=a^2-c^2$ and hence $c^2=a^2-b^2$ .

Put $c^2=a^2-b^2$ in the above expression to get,

  $\begin{array}{c}{x}^2+y^2+a^2-b^2=\frac{\left(a^2-b^2\right)y^2}{a^2}+a^2\\ x^2+y^2-b^2=\left(1-\frac{b^2}{a^2}\right)y^2\\ x^2+y^2-b^2=y^2-\frac{b^2}{a^2}{y}^2\\ x^2-b^2=-\frac{b^2}{a^2}{y}^2\end{array}$

Simplifying,

  $\begin{array}{c}{x}^2-b^2=-\frac{b^2}{a^2}{y}^2\\ x^2+\frac{b^2}{a^2}{y}^2=b^2\\ \frac{x^2}{b^2}+\frac{y^2}{a^2}=1\end{array}$

Equivalently,

  $\frac{y^2}{a^2}+\frac{x^2}{b^2}=1$

Thus, it has been shown that an equation for the ellipse with the given center, the given foci, and the given semi-major axis is $\frac{y^2}{a^2}+\frac{x^2}{b^2}=1$ , where $b^2=a^2-c^2$ .

Conclusion:

It has been shown that an equation for the ellipse with the given center, the given foci, and the given semi-major axis is $\frac{y^2}{a^2}+\frac{x^2}{b^2}=1$ , where $b^2=a^2-c^2$ .