Chapter 9.2, Problem 55E

To determine

To find: The standard form of the equation of ellipse with center at origin and major axis on the x-axis with given conditions.

Answer to Problem 55E

The equation of ellipse is $\frac{x^2}{{\left(17.835\right)}^2}+\frac{y^2}{{\left(4.34\right)}^2}=1$ .

Explanation of Solution

Given information:

Halley’s comet has an elliptical orbit with sun at one focus. The eccentricity of the orbit is approximately $0.97$ . The length of the major axis is about $35.67$ astronomical units.

Calculation:

The major axis of the ellipse is horizontal.

So, the general equation of ellipse is $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$ .

Use the formula for length of the major axis.

  $\begin{array}{l}2a=35.67\\ a=\frac{35.67}{2}\\ a=17.835\end{array}$

So, the value of $a$ is equal to $17.835$ astronomical units.

Use the formula for eccentricity of ellipse.

  $\begin{array}{c}e=\frac{c}{a}\\ 0.97=\frac{c}{17.835}\\ c=17.835\times 0.97\\ c=17.30\end{array}$

Use the formula.

  $\begin{array}{c}{c}^2=a^2-b^2\\ {\left(17.3\right)}^2={\left(17.835\right)}^2-b^2\\ b^2=318.09-299.29\\ b^2=18.8\\ b\approx 4.34\end{array}$

Now substitute $17.835$ for $a$ and $4.34$ for $b$ , $0$ for $h$ and $0$ for $k$ in the general equation of ellipse.

  $\begin{array}{c}\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1\\ \frac{{\left(x-0\right)}^2}{{\left(17.835\right)}^2}+\frac{{\left(y-0\right)}^2}{{\left(4.34\right)}^2}=1\\ \frac{x^2}{{\left(17.835\right)}^2}+\frac{y^2}{{\left(4.34\right)}^2}=1\end{array}$

Therefore, the equation of ellipse is $\frac{x^2}{{\left(17.835\right)}^2}+\frac{y^2}{{\left(4.34\right)}^2}=1$ .