Question

Halley’s comet has an elliptical orbit with the sun at one focus. The eccentricity of the orbit is approximately 0.97. The length of the major axis of the orbit is about 35.88 astronomical units. (An astronomical unit is about 93 million miles.) Find the standard form of the equation of the orbit. Place the center of the orbit at the origin and place the major axis on the x-axis.

Answer 1

so the sun is one of the foci, and we have an eccentricity of 0.97. So the ellipse will look more or less like the picture below.

$\bf \textit{ellipse, horizontal major axis}\\\\\cfrac{(x- h)^2}{ a^2}+\cfrac{(y- k)^2}{ b^2}=1\qquad \begin{cases}center\ ( h, k)\\vertices\ ( h\pm a, k)\\c=\textit{distance from}\\\qquad \textit{center to foci}\\\qquad \sqrt{ a ^2- b ^2}\\eccentricity\quad e=\cfrac{c}{a}\end{cases}\\\\-------------------------------\\\\\begin{cases}h=0\\k=0\\a=35.88\\e=0.97\end{cases}\implies \cfrac{(x-0)^2}{35.88^2}+\cfrac{(y-0)^2}{b^2}=1\\\\-------------------------------$

$\bf e=\cfrac{c}{a}\implies 0.97=\cfrac{c}{35.88}\implies \boxed{34.8036=c} \\\\\\ c=\sqrt{a^2-b^2}\implies b=\sqrt{a^2-c^2}\implies b=\sqrt{35.88^2-34.8036^2} \\\\\\ \boxed{b\approx 8.72}\\\\ -------------------------------\\\\ \cfrac{(x-0)^2}{35.88^2}+\cfrac{(y-0)^2}{8.72^2}=1\implies \cfrac{x^2}{1287.3744}+\cfrac{y^2}{76.0384}=1$