Planet M orbits around its sun,S,in an ellipt | vedclass

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Question

(A) Let the major axis be along the $y$-axis. The distance between the closest point $M_1$ and farthest point $M_2$ is the length of the major axis $2

Vedclass · Accepted Answer

$\frac{x^2}{36} + \frac{(y+8)^2}{100} = 1$

Vedclass · Answer

(A) Let the major axis be along the $y$-axis. The distance between the closest point $M_1$ and farthest point $M_2$ is the length of the major axis $2b = 18 + 2 = 20$,so $b = 10$.Given the sun $S$ is at $(0, 0)$ and is a focus,the distance from the center $(0, -k)$ to the focus is $be = 8$. Since the focus $S$ is at $(0, 0)$ and the other focus is on the negative $y$-axis,the center must be at $(0, -8)$.Thus,$b = 10$ and $be = 8$,which implies $e = 0.8$.Using $a^2 = b^2(1 - e^2)$,we get $a^2 = 100(1 - 0.64) = 100(0.36) = 36$,so $a = 6$.The equation of the ellipse with center $(0, -8)$ and major axis along the $y$-axis is $\frac{x^2}{a^2} + \frac{(y - (-8))^2}{b^2} = 1$,which simplifies to $\frac{x^2}{36} + \frac{(y+8)^2}{100} = 1$.

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