Problem 58 - College Algebra

Question

For the following exercises, find the area of the ellipse. The area of an ellipse is given by the formula Area \(=a \cdot b \cdot \pi\). $$ \frac{(x+6)^{2}}{16}+\frac{(y-6)^{2}}{36}=1 $$

Step-by-Step Solution

Answer

The area of the ellipse is \(24\pi\).

Step 1: Identify standard form of the ellipse equation

The given ellipse equation is \(\frac{(x+6)^{2}}{16}+\frac{(y-6)^{2}}{36}=1\). This matches the standard form of an ellipse equation \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\) with \(h = -6\), \(k = 6\), \(a^2 = 16\), and \(b^2 = 36\).

Step 2: Calculate the semi-major and semi-minor axes

From \(a^2 = 16\), we find \(a = \sqrt{16} = 4\).From \(b^2 = 36\), we find \(b = \sqrt{36} = 6\).

Step 3: Compute the area of the ellipse

The formula for the area of an ellipse is given by \(\text{Area} = a \cdot b \cdot \pi\). Substitute \(a = 4\) and \(b = 6\) into the formula:\(\text{Area} = 4 \cdot 6 \cdot \pi = 24\pi\).

Key Concepts

Standard Form of Ellipse

The equation of an ellipse can be expressed in a standard form, which helps us easily identify its properties. The standard form is written as:

• \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \)

This formula reveals the center of the ellipse at the point \((h, k)\). It visually represents how the ellipse is shifted in the Cartesian plane. The values of \(a^2\) and \(b^2\) denote the squares of the lengths of the semi-major and semi-minor axes, respectively.

For example, in the equation \( \frac{(x+6)^2}{16} + \frac{(y-6)^2}{36} = 1 \), the center of the ellipse is at \((-6, 6)\). The numbers under the squared terms, here 16 and 36, help us calculate the lengths of the semi-major and semi-minor axes.

Semi-Major Axis

The semi-major axis is one of the two defining radii of an ellipse. It is the longest radius, stretching from the center to the furthest point on the ellipse in the direction parallel to the major axis.

In the standard form equation, \(a^2\) is the term associated with the semi-major axis when \(a^2 > b^2\). For the equation \( \frac{(x+6)^2}{16} + \frac{(y-6)^2}{36} = 1 \), the term \(36\) represents \(b^2\). Hence, the semi-major axis is along the y-direction and its length \(b\) is calculated as:

• \(b = \sqrt{36} = 6 \)

This indicates that the length of the semi-major axis is 6 units.

Semi-Minor Axis

Like the semi-major axis, the semi-minor axis is a crucial part of the ellipse. It is the shortest radius stretching from the ellipse's center to the edge, perpendicular to the major axis.

In the standard ellipse equation, \(b^2\) corresponds to the semi-minor axis if \(b^2 < a^2\). In our given ellipse equation, the term \(16\) is \(a^2\). Thus, the semi-minor axis aligns with the x-direction, and its length \(a\) is:

• \(a = \sqrt{16} = 4 \)

Thus, the semi-minor axis of the ellipse is 4 units long.