Problem 61 - Intermediate Algebra

Question

The area \(A\) bounded by an ellipse with the equation \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is given by \(A=\pi a b\). Find the area bounded by the ellipse described by \(9 x^{2}+16 y^{2}=144\).

Step-by-Step Solution

Answer

The area bounded by the ellipse is \(12\pi\).

Step 1: Identify the Standard Form

The given equation is \(9x^2 + 16y^2 = 144\). Divide each term by 144 to convert it into the standard form of an ellipse equation.\[\frac{x^2}{16} + \frac{y^2}{9} = 1\]

Step 2: Relate the Form to Parameters

The standard form of the ellipse is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). By comparing, we find \(a^2 = 16\) and \(b^2 = 9\). Thus, \(a = 4\) and \(b = 3\).

Step 3: Calculate the Area

Use the formula \(A = \pi a b\) to find the area of the ellipse. Substitute \(a = 4\) and \(b = 3\) into the formula.\[A = \pi \times 4 \times 3 = 12\pi\]

Step 4: Conclusion

The area bounded by the ellipse \(9x^2 + 16y^2 = 144\) is \(12\pi\).

Key Concepts

Standard form of ellipse

We start with understanding the standard form of an ellipse, which is crucial for identifying the parameters that define its size and shape. An ellipse is essentially an elongated circle, and its standard form equation is given by:\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]Here, \(a\) and \(b\) are the length of the semi-major and semi-minor axes, respectively. By observing this equation, you can immediately recognize the proportions and orientation of the ellipse. Depending on whether \(a > b\) or \(b > a\), the ellipse will be oriented either horizontally or vertically. Breaking it down, this form allows us to easily see the dimensions of the ellipse:

• \(a^2\) under \(x^2\) represents the horizontal stretch.
• \(b^2\) under \(y^2\) represents the vertical stretch.

This form helps in conversion from any general quadratic form by dividing each term in the equation with the constant on the right-hand side to achieve '1' on the right.

Equation of ellipse

When dealing with the equation of an ellipse, knowing how to manipulate and understand it is essential. Given any ellipse equation:\[Ax^2 + By^2 = C\]you can convert it into the standard form by dividing through by \(C\). For instance, take the equation \[9x^2 + 16y^2 = 144\].Divide the entire equation by 144:\[\frac{x^2}{16} + \frac{y^2}{9} = 1\]This conversion is straightforward but important as it sets up the ellipse in a standardized way, making it easier to interpret. In this particular case, it's easy to observe that the semi-major axis is larger than the semi-minor axis, indicating a horizontal ellipse.Comparing with the standard form, we deduce:

• \(a^2 = 16\), hence \(a = 4\)
• \(b^2 = 9\), hence \(b = 3\)

By extracting \(a\) and \(b\) from the equation, we can easily move to step calculating the area.

Calculating area of ellipse

Once we have the parameters of the ellipse, calculating its area becomes a simple task. The formula to find the area \(A\) of an ellipse is given by:\[A = \pi a b\]where \(a\) is the semi-major axis, and \(b\) is the semi-minor axis. Let's apply this understanding to the ellipse in our example, where we found \(a = 4\) and \(b = 3\).Simply substitute these values into the formula:\[A = \pi \times 4 \times 3 = 12\pi\]This shows that the area of the ellipse, bounded by the equation \(9x^2 + 16y^2 = 144\), is \(12\pi\). It's a neat reflection of how geometry and algebra come together to solve practical problems. This calculation is straightforward once the ellipse is in its standard form, emphasizing the efficiency of organizing your approach to these kinds of problems.