Question

# Comet Orbit Halley’s comet has an elliptical orbit, with the sun at one focus. The eccentricity of the orbit is approximately $0.967$. The length of the major axis of the orbit is approximately $35.88$ astronomical units. (An astronomical unit is about $93$ million miles.)Find an equation of the orbit. Place the center of the orbit at the origin, and place the major axis on the $x$-axis.

Solution

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Given:

\begin{aligned} e&=\text{Eccentricity}=0.967 \\ 2a&=\text{Length major axis}=35.88\text{ astronomical units} \end{aligned}

We need to determine an equation for the ellipse, assuming that the origin corresponds with the center and the major axis is horizontal.

Standard form of the equation of an ellipse ($0):

\begin{aligned} \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}&=1 &\textcolor{#4559AC}{\text{Horizontal major axis}} \\ \frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}&=1 &\textcolor{#4559AC}{\text{Vertical major axis}} \end{aligned}

• The length of the major axis is $2a$.
• Center is $(h,k)$
• Eccentricity: $e=\frac{c}{a}$ with $c^2=a^2-b^2$

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