Solve the problem.
-The orbit of a planet around a sun is an ellipse with the sun at one focus. The aphelion of a planet is its greatest distance from the sun, its perihelion is its shortest distance, and its mean distance is the length of the semimajor
Axis of the elliptical orbit. If a planet has a perihelion of 402.3 million miles and a mean distance of 404 million miles, write an equation for the orbit of the planet around the sun.
A) $\frac { x ^ { 2 } } { 404 ^ { 2 } } + \frac { y ^ { 2 } } { 403.996 ^ { 2 } } = 1$
B) $\frac { x ^ { 2 } } { 404 ^ { 2 } } + \frac { y ^ { 2 } } { 1.7 ^ { 2 } } = 1$
C) $\frac { x ^ { 2 } } { 404.004 ^ { 2 } } + \frac { y ^ { 2 } } { 404 ^ { 2 } } = 1$
D) $\frac { x ^ { 2 } } { 404 ^ { 2 } } + \frac { y ^ { 2 } } { 402.3 ^ { 2 } } = 1$

Question

Solve the problem.
-The orbit of a planet around a sun is an ellipse with the sun at one focus. The aphelion of a planet is its greatest distance from the sun, its perihelion is its shortest distance, and its mean distance is the length of the semimajor
Axis of the elliptical orbit. If a planet has a perihelion of 402.3 million miles and a mean distance of 404 million miles, write an equation for the orbit of the planet around the sun. 
A) $\frac { x ^ { 2 } } { 404 ^ { 2 } } + \frac { y ^ { 2 } } { 403.996 ^ { 2 } } = 1$
B) $\frac { x ^ { 2 } } { 404 ^ { 2 } } + \frac { y ^ { 2 } } { 1.7 ^ { 2 } } = 1$
C) $\frac { x ^ { 2 } } { 404.004 ^ { 2 } } + \frac { y ^ { 2 } } { 404 ^ { 2 } } = 1$
D) $\frac { x ^ { 2 } } { 404 ^ { 2 } } + \frac { y ^ { 2 } } { 402.3 ^ { 2 } } = 1$

Quizplus · Accepted Answer

The equation of an ellipse with the sun at one focus is given by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a$ is the semi-major axis and $b$ is the semi-minor axis. The mean distance, which is the length of the semi-major axis, is given as 404 million miles, so $a = 404$. To find $b$, we use the fact that the distance from the center to a focus, $c$, is given by $c = a - \text{perihelion} = 404 - 402.3 = 1.7$ million miles. Since $b^2 = a^2 - c^2$, we find $b^2 = 404^2 - 1.7^2$, which gives a semi-minor axis slightly less than 404, making option A the correct choice because it correctly represents the dimensions of the ellipse with the given conditions.

Solve the Problem x24042+y2403.9962=1\frac { x ^ { 2 } } { 404 ^ { 2 } } + \frac { y ^ { 2 } } { 403.996 ^ { 2 } } = 14042x2​+403.9962y2​=1

Solve the Problem $\frac { x ^ { 2 } } { 404 ^ { 2 } } + \frac { y ^ { 2 } } { 403.996 ^ { 2 } } = 1$