A planet orbits a star along an elliptical path, with the star at a focus of the ellipse. The planet's path can be modeled by the ellipse shown below, centered at the origin, with a horizontal major axis. If the minimum distance from the planet to the star is 98 million km and the maximum distance is 200 million km, write the equation of the ellipse (in millions of km).


To find the equation of the ellipse, let's use the information provided. The standard form of the equation of an ellipse with a horizontal major axis and centered at the origin is:

\[
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
\]

Where:
- \( a \) is the semi-major axis (half the length of the major axis),
- \( b \) is the semi-minor axis (half the length of the minor axis),
- The foci are located at \( (\pm c, 0) \), where \( c \) is the distance from the center to a focus.

### Step 1: Determine the values of \( a \) and \( c \)
- The minimum distance from the planet to the star (focus) is 98 million km, and this corresponds to the closest point on the ellipse, which is at one of the vertices. The vertex is  \( a - c = 98 \).
- The maximum distance from the planet to the star is 200 million km, which corresponds to the farthest point, so \( a + c = 200 \).

We now have two equations:
1. \( a - c = 98 \)
2. \( a + c = 200 \)

Solving these two equations together:
- Add the two equations: 
  \[
  (a - c) + (a + c) = 98 + 200 \implies 2a = 298 \implies a = 149
  \]
- Now substitute \( a = 149 \) into either equation, say \( a - c = 98 \):
  \[
  149 - c = 98 \implies c = 51
  \]

### Step 2: Find \( b \)
Using the relationship between \( a \), \( b \), and \( c \) for an ellipse, which is:
\[
c^2 = a^2 - b^2
\]
Substitute \( a = 149 \) and \( c = 51 \):
\[
51^2 = 149^2 - b^2 \implies 2601 = 22201 - b^2 \implies b^2 = 22201 - 2601 = 19600 \implies b = \sqrt{19600} = 140
\]

### Step 3: Write the equation of the ellipse
Now that we have \( a = 149 \) and \( b = 140 \), we can write the equation of the ellipse:
\[
\frac{x^2}{149^2} + \frac{y^2}{140^2} = 1
\]
Simplifying the squares:
\[
\frac{x^2}{22201} + \frac{y^2}{19600} = 1
\]

This is the equation of the ellipse in millions of km.

Would you like more details on any part of the solution?

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Here are some related questions:
1. How do you find the foci of an ellipse?
2. What is the relationship between the semi-major and semi-minor axes?
3. Can an ellipse have vertical major axes?
4. What does the distance between the foci tell us about the shape of the ellipse?
5. How would the equation change if the ellipse were centered elsewhere?

**Tip:** In an ellipse, the total distance from any point on the ellipse to the two foci is constant.

A planet orbits a star along an elliptical path, with the star at a focus of the ellipse. The planet's path can be modeled by the ellipse shown below, centered at the origin, with a horizontal major axis. If the minimum distance from the planet to the star is 98 million km and the maximum distance is 200 million km, write the equation of the ellipse (in millions of km).

Learn how to find the equation of an ellipse that models the path of a planet orbiting a star. Given the minimum distance of 98 million km and maximum distance of 200 million km, this problem covers key geometric concepts of ellipses and includes a step-by-step solution. Suitable for students in Grades 10-12.

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