Problem 19 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 19-30: Find an equation for the ellipse that has its center at the origin and satisfies the given conditions. Vertices \(V(\pm 8,0)\), $$ \text { foci } F(\pm 5,0) $$

Step-by-Step Solution

Answer

The equation of the ellipse is \( \frac{x^2}{64} + \frac{y^2}{39} = 1 \).

Step 1: Identify the Ellipse Type

Since the vertices and foci are on the x-axis, this is a horizontally-oriented ellipse.

Step 2: Write the General Equation

The general equation for a horizontal ellipse centered at the origin is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( a \) is the semi-major axis length and \( b \) is the semi-minor axis length.

Step 3: Determine the Values of \( a \) and \( c \)

The vertices \( V(\pm 8, 0) \) indicate that \( a = 8 \), and the foci \( F(\pm 5, 0) \) provide \( c = 5 \).

Step 4: Calculate \( b \) Using the Relationship \( c^2 = a^2 - b^2 \)

Use the relationship \( c^2 = a^2 - b^2 \). Given \( c = 5 \), \( a = 8 \), we have:\[5^2 = 8^2 - b^2 \25 = 64 - b^2 \b^2 = 64 - 25 = 39 \b = \sqrt{39}\]

Step 5: Plug \( a^2 \) and \( b^2 \) Into the Equation

Replace \( a^2 \) with \( 64 \) and \( b^2 \) with \( 39 \) in the general equation:\[ \frac{x^2}{64} + \frac{y^2}{39} = 1 \]

Key Concepts

Semi-Major Axis

In an ellipse, the semi-major axis is the longest radius from the center to the edge. It defines how "stretched out" the ellipse is. For a horizontally oriented ellipse, this axis runs along the x-axis. In the given exercise, the vertices are at

• \( V(\pm 8, 0) \)

indicating that the semi-major axis is 8 units long. This means that from the center (which is at the origin, (0,0)), the ellipse extends 8 units to the left and 8 units to the right. The length of the semi-major axis is

• \( a = 8 \).

Semi-Minor Axis

The semi-minor axis of an ellipse is the shortest radius from the center to the edge. For a horizontal ellipse, this axis runs along the y-axis. In our problem, we need to calculate the semi-minor axis using the relationship between the semi-major axis, the

• foci
• and the semi-minor axis known as:

\( c^2 = a^2 - b^2 \).Here, "c" is the distance from the center to each focus. Given that the foci are at

• \( F(\pm 5, 0) \)

we have

• \( c = 5 \).

Using

• \( a = 8 \)

we solve to find

• \( b^2 = 64 - 25 = 39 \).

Thus,

• \( b = \sqrt{39} \).

Foci

The foci (singular: focus) of an ellipse are two fixed points located inside the ellipse. They have an important role as they define the shape and size of the ellipse. The sum of the distances from any point on the ellipse to the two foci is constant. For our particular ellipse, the foci are at

• \( F(\pm 5, 0) \)

meaning they are 5 units away from the center along the x-axis. The presence of foci is a critical part of accurately describing an ellipse's geometry, as it helps differentiate it from a circle.

Ellipse Equation

The ellipse equation serves as a precise mathematical description of the ellipse’s shape, allowing one to easily graph it or find specific points. For a horizontal ellipse centered at the origin, the standard form of the equation is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).In this case, the calculated values are used:

• The semi-major axis length (\(a\)) is 8, thus \(a^2 = 64\), and we place this value under the \(x^2\) term.
• The semi-minor axis length (\(b\)) is \(\sqrt{39}\), leading to \(b^2 = 39\), which we place under the \(y^2\) term.

So the ellipse equation becomes:

• \[ \frac{x^2}{64} + \frac{y^2}{39} = 1 \]

This equation allows us to accurately plot or understand any attributes of the ellipse on a coordinate plane.