Problem 22 - Precalculus with Calculus Previews

Question

Find an equation of the ellipse that satisfies the given conditions. Vertices \((\pm 9,0),\) foci (±2,0)

Step-by-Step Solution

Answer

The equation of the ellipse is \(\frac{x^2}{81} + \frac{y^2}{77} = 1\).

Step 1: Identify the Orientation of the Ellipse

The vertices and foci along the x-axis suggest that the ellipse is aligned horizontally. This indicates the general form of the ellipse's equation will be \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), with \(a > b\).

Step 2: Establish the Values of 'a' and 'c'

From the given vertices \((\pm 9, 0)\), we identify \( a = 9 \) because the distance from the center to a vertex on the horizontal axis is 9. From the foci \((\pm 2, 0)\), we identify \( c = 2 \) since the distance from the center to a focus is 2.

Step 3: Calculate the Value of 'b' Using the Formula for Ellipses

In ellipses where \(a > b\), we use the relation \( c^2 = a^2 - b^2 \) to find \(b\). Substituting the known values, we have \[c^2 = a^2 - b^2\] \[2^2 = 9^2 - b^2\] \[4 = 81 - b^2\]. Solving for \(b^2\), we get \(b^2 = 77\).

Step 4: Write the Equation of the Ellipse

Substituting the values of \(a\) and \(b\) into the general equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), we obtain: \[\frac{x^2}{81} + \frac{y^2}{77} = 1\]. This is the equation of the ellipse.

Key Concepts

Ellipse Vertices

The vertices of an ellipse are crucial points that help define its shape and size. They are the furthest points along the ellipse's principal axis from the center. In this exercise, the vertices \((\pm 9, 0)\) indicate that the ellipse is placed horizontally along the x-axis.

• The vertices are located symmetrically around the origin, implying that the center of the ellipse is at \((0,0)\).
• The distance from the center to each vertex, denoted as \(a\), determines the semi-major axis of the ellipse.

In our specific case, the vertices are at \((\pm 9, 0)\), meaning the semi-major axis is 9 units long. Thus, \(a = 9\).
Knowing the vertices allows us to determine the maximum width of the ellipse, which is \(2a = 18\) units.

Ellipse Foci

Foci are special points inside an ellipse used to define its elliptical shape. These points are located along the major axis, and the total distance from one focus to a point on the ellipse and back to the other focus remains constant.

• In this problem, the foci are given as \((\pm 2, 0)\). This tells us that they lie on the major axis, symmetrically around the center.
• The distance from the center to each focus is denoted as \(c\).

For our ellipse, \(c = 2\), indicating the foci lie 2 units from the center on the x-axis.
Using the relationship \(c^2 = a^2 - b^2\), a feature of ellipses, we can explore the relationship between the axes and foci further.

Ellipse Equation Derivation

The equation of an ellipse is derived using the values of the semi-major axis, the semi-minor axis, and the foci. For a horizontally aligned ellipse, the general form of its equation is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\).
Steps to derive the equation:

• Verify the orientation of the ellipse based on its vertices and foci placement.
• Identify \(a = 9\) from the vertices \((\pm 9, 0)\).
• Identify \(c = 2\) from the foci \((\pm 2, 0)\).
• Using \(c^2 = a^2 - b^2\), calculate \(b\). With \(a = 9\) and \(c = 2\), we substitute into \(4 = 81 - b^2\).

Solving gives \(b^2 = 77\), resulting in the semi-minor axis length.Finally, substitute \(a^2 = 81\) and \(b^2 = 77\) into the general ellipse equation:\[\frac{x^2}{81} + \frac{y^2}{77} = 1\], giving us a complete representation of the ellipse in equation form.