Chapter 4: Problem 42
Find a positive number \(c\) such that the area inside the ellipse $$ c x^{2}+7 y^{2}=3 $$ is 2
Short Answer
Expert verified
The positive number c that makes the area inside the ellipse equal to 2 is \(c = \frac{9Ï€^2}{28}\).
Step by step solution
01
Rewrite the ellipse equation in standard form
To find the values of a and b, first rewrite the ellipse equation in standard form. The standard form of an ellipse equation is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where a and b are the lengths of the semi-major and semi-minor axes. Divide both sides of the given equation by 3:
\[\frac{cx^2}{3} + \frac{7y^2}{3} = 1\]
02
Express a and b in terms of c
From the standard form, we can express a and b in terms of c as follows:
\(a^2 = \frac{3}{c}\) and \(b^2 = \frac{3}{7}\).
Since a and b are positive, we can take the square root of these equations:
\(a = \sqrt{\frac{3}{c}}\) and \(b = \sqrt{\frac{3}{7}}\).
03
Use the formula for the area of an ellipse
We know that the area inside the ellipse is 2. Using the formula for the area of an ellipse: \(Area = πab\), we have:
\(2 = π \cdot \sqrt{\frac{3}{c}} \cdot \sqrt{\frac{3}{7}}\)
04
Solve for c
Now, we can solve the above equation for c:
\(2 = \frac{3Ï€}{\sqrt{7c}}\),
Multiply both sides by the denominator:
\(\sqrt{7c} = \frac{3Ï€}{2}\),
Now, square both sides to get rid of the square root:
\(7c = \left(\frac{3Ï€}{2}\right)^2\),
Divide both sides by 7:
\(c = \frac{\left(\frac{3Ï€}{2}\right)^2}{7}\),
Simplify the equation:
\(c = \frac{9Ï€^2}{28}\).
So, the positive number c that makes the area inside the ellipse equal to 2 is \(c = \frac{9Ï€^2}{28}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Standard Form of an Ellipse
The standard form of an ellipse is a fundamental way of expressing the ellipse equation. It helps us describe the shape and properties of an ellipse in a simple form. The standard equation is given by \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] where:
- \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes respectively.
- The equation is normalized to equal 1, setting a clear representation of the ellipse based on the axes' lengths.
Area of an Ellipse
The area of an ellipse is calculated using a simple formula closely related to the lengths of its axes. The formula is \[ \text{Area} = \pi ab \] where:
- \(a\) is the length of the semi-major axis.
- \(b\) is the length of the semi-minor axis.
- \(\pi\) (pi) is a constant roughly equal to 3.14159.
Semi-Major and Semi-Minor Axes
Ellipses have two principal axes: the semi-major and semi-minor axes. Understanding their roles is crucial to working with ellipses effectively.
- The **semi-major axis** (\(a\)) is the longest diameter of the ellipse. It stretches from one end of the ellipse to the opposite end, passing through the center.
- The **semi-minor axis** (\(b\)) is perpendicular to the semi-major axis at the center and is the shortest diameter.