Problem 59 - College Algebra

Question

For the following exercises, find the area of the ellipse. The area of an ellipse is given by the formula Area \(=a \cdot b \cdot \pi\). $$ \frac{(x+1)^{2}}{4}+\frac{(y-2)^{2}}{5}=1 $$

Step-by-Step Solution

Answer

The area of the ellipse is approximately \( 14.031\pi \).

Step 1: Identify the Equation of the Ellipse

The given equation \( \frac{(x+1)^{2}}{4} + \frac{(y-2)^{2}}{5} = 1 \) is in the standard form \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \). Here, \( h = -1 \), \( k = 2 \), \( a^2 = 4 \), and \( b^2 = 5 \).

Step 2: Determine the Lengths of the Semi-Axes

Find \( a \) and \( b \) by taking the square roots of \( a^2 \) and \( b^2 \). Thus, \( a = \sqrt{4} = 2 \) and \( b = \sqrt{5} \approx 2.236 \).

Step 3: Apply the Formula for the Area of an Ellipse

Use the formula for the area of an ellipse, \( \text{Area} = \pi \cdot a \cdot b \). Substitute the values of \( a \) and \( b \): \[ \text{Area} = \pi \cdot 2 \cdot \sqrt{5} \approx \pi \cdot 2 \cdot 2.236 \approx \pi \cdot 4.472. \]

Step 4: Compute the Approximate Area

Finally, calculate the area using the approximation for \( b \): \( \text{Area} \approx 14.031 \pi \).

Key Concepts

Standard Form of Ellipse Equation

An ellipse is a smooth, closed shape that looks like an elongated circle. To understand and work with ellipses, especially to calculate their area, it's essential to know their standard form equation. The most common representation is:\[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]Here’s what each component represents:

• \(h\) and \(k\) are the coordinates of the center of the ellipse. In simple terms, they tell you where the center of the ellipse is located on the graph.
• \(a^2\) and \(b^2\) are the denominators that help define the shape's size. \(a^2\) is associated with the horizontal radius, and \(b^2\) with the vertical radius.

In the equation given in the exercise, \(h = -1\) and \(k = 2\) indicate the ellipse is centered at (-1, 2) on the coordinate plane.
Identifying these components helps set the base for solving area and knowing more about the ellipse’s dimensions.

Semi-Axes of Ellipse

The semi-axes are semi-major and semi-minor axes, essentially the ellipse's 'radii.' These are the maximum and minimum distances from the center to the edge of the ellipse.

• The semi-major axis is the longer radius, which corresponds to the larger denominator in the standard form equation. It can be horizontal or vertical, depending on which is larger, \(a^2\) or \(b^2\).
• The semi-minor axis is the shorter radius. Similarly, this is associated with the smaller denominator.

In our example, \(a^2 = 4\) and \(b^2 = 5\). To find the actual lengths of the semi-axes, calculate their square roots:

• \(a = \sqrt{4} = 2\)
• \(b = \sqrt{5} \approx 2.236\)

Knowing these lengths is key to determining the area of the ellipse and understanding its geometry.

Formula for Area of Ellipse

The formula to find the area of an ellipse is straightforward once you know the lengths of the semi-major and semi-minor axes:\[\text{Area} = \pi \cdot a \cdot b\]Here's how it works step-by-step:

• The number \(\pi\) is a constant you are likely familiar with from circle calculations, approximately equal to 3.14159.
• Multiply \(\pi\) by the product of the lengths of the semi-axes \(a\) and \(b\).

For our ellipse where \(a = 2\) and \(b = \sqrt{5}\), the area is calculated as:\[\text{Area} = \pi \cdot 2 \cdot \sqrt{5}\]This simple calculation provides the total enclosed area of the ellipse.

Approximation of Square Root

Approximating square roots is a helpful skill in geometry and in finding quick values for calculations like the ellipse's area. Here's a brief guide on how to approximate:

• Identify the perfect squares closest to your number. For example, \(4\) and \(9\) are perfect squares near \(5\).
• \(\sqrt{4} = 2\) and \(\sqrt{9} = 3\). Since \(5\) is close to \(4\), \(\sqrt{5}\) is a little over \(2\).

In practice, for quick calculations or when a calculator isn’t handy, you might call \(\sqrt{5} \approx 2.236\) by comparing it with known values and adjusting slightly.Using this approximation helps simplify problems involving non-perfect squares, allowing for easier, faster calculations, which is especially practical for problems like calculating the ellipse's area.