Problem 57 - College Algebra

Question

For the following exercises, find the ellipse. The area of an ellipse is given by the formula Area \(=a \cdot b \cdot \pi\) $$ \frac{(x-3)^{2}}{9}+\frac{(y-3)^{2}}{16}=1 $$

Step-by-Step Solution

Answer

The area of the ellipse is \(12\pi\).

Step 1: Identify the Ellipse Formula

The given equation is \( \frac{(x-3)^2}{9} + \frac{(y-3)^2}{16} = 1 \). This is in the standard form of an ellipse equation, \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), where \((h, k)\) is the center, and \(a\) and \(b\) are the semi-major and semi-minor axes, respectively.

Step 2: Extract Values for a and b

From the equation \( \frac{(x-3)^2}{9} + \frac{(y-3)^2}{16} = 1 \), we can identify \( a^2 = 9 \) and \( b^2 = 16 \). Therefore, \( a = 3 \) and \( b = 4 \).

Step 3: Calculate the Area of the Ellipse

The area of an ellipse is given by the formula \( \text{Area} = a \cdot b \cdot \pi \). Substituting \( a = 3 \) and \( b = 4 \), we calculate the area as:\[ \text{Area} = 3 \cdot 4 \cdot \pi = 12\pi \]

Key Concepts

Ellipse Equation

The general equation for an ellipse is \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]where

• \((h, k)\) represents the center of the ellipse.
• \(a\) and \(b\) are constants that determine the dimensions of the ellipse.

In the provided equation, \(\frac{(x-3)^2}{9} + \frac{(y-3)^2}{16} = 1\), the center is located at \((h, k) = (3, 3)\). It shows us that every point \((x, y)\) on the ellipse satisfies this equation. This form is helpful because it reveals crucial properties like the lengths of the axes and the position of the ellipse on a coordinate plane.When you are working with ellipses, always ensure your equation matches this standard form. This will help you to easily identify the center and the lengths of the semi-major and semi-minor axes.

Semi-Major Axis

The semi-major axis, denoted by \(a\), is the longest radius of an ellipse. In the standard form ellipse equation, it's represented as the denominator where the variable change is minor but along the axis with larger spread.From \[\frac{(x-3)^2}{9} + \frac{(y-3)^2}{16} = 1\]we identify \(b^2 = 16\), making \(b\) with the larger term equal to 4. This tells us the length of the semi-major axis of the ellipse. If \(b > a\), the semi-major axis is along the y-direction.The length of the full major axis is \(2b\); thus, it is \(8\) in this context. This is a characteristic of all ellipses, defining how stretched it is along its primary direction.

Semi-Minor Axis

The semi-minor axis is the shorter radius of the ellipse. Using the standard ellipse equation form, this is the smaller value between \(a\) and \(b\) in the denominators. In the exercise equation \[\frac{(x-3)^2}{9} + \frac{(y-3)^2}{16} = 1\]we identify \(a^2 = 9\). Solving for \(a\) (where \(a = \sqrt{9}\)) gives us a semi-minor axis length of \(3\).The full minor axis, therefore, measures \(6\) units wide.

• This axis affects how compressed the ellipse looks compared to its semi-major axis.
• It also helps you understand the overall structure of the ellipse.

Understanding both axes allows you to visualize the ellipse in its coordinate space.

Area of an Ellipse

The area of an ellipse is calculated using the formula:\[\text{Area} = a \cdot b \cdot \pi\]This formula uses the lengths of the semi-major \(b\) and semi-minor \(a\) axes.In our example, the values are \(a = 3\) and \(b = 4\). Thus, substituting these into the formula gives:\[\text{Area} = 3 \cdot 4 \cdot \pi = 12\pi\]The presence of \(\pi\) in the area formula shows the relationship of ellipses to circles, where the circumference and area are also expressed using \(\pi\). This result reflects how much space the ellipse covers in its plane.Understanding this formula is crucial as it links geometry's abstract concepts to real-world shapes, offering practical insights.