The orbit of Jupiter around the sun is an ellipse with equation $\frac { x ^ { 2 } } { 6561 } + \frac { y ^ { 2 } } { 5329 } = 1$ where x and y are measured in millions of miles. Find the eccentricity of this ellipse. (Round to the nearest thousandth.)
A) 0.812
B) 15.210
C) 1.231
D) 0.433

Question

Quizplus · Accepted Answer

The eccentricity of an ellipse is given by the formula $e = \sqrt{1 - \frac{b^2}{a^2}}$, where $a$ and $b$ are the semi-major and semi-minor axes, respectively. In the given equation, $a^2 = 6561$ and $b^2 = 5329$, so $a = 81$ and $b = 73$. Plugging these into the formula gives $e = \sqrt{1 - \frac{5329}{6561}} = \sqrt{1 - \frac{73^2}{81^2}} = \sqrt{1 - \frac{5329}{6561}} = \sqrt{\frac{1232}{6561}}$, which when calculated gives an eccentricity of approximately 0.433.

The Orbit of Jupiter Around the Sun Is an Ellipse x26561+y25329=1\frac { x ^ { 2 } } { 6561 } + \frac { y ^ { 2 } } { 5329 } = 16561x2​+5329y2​=1

The Orbit of Jupiter Around the Sun Is an Ellipse $\frac { x ^ { 2 } } { 6561 } + \frac { y ^ { 2 } } { 5329 } = 1$