Problem 28 - Algebra for College Students

Question

Find an equation of the ellipse that satisfies the oiven conditions. Vertices \((\pm 4,0)\), foci \((\pm 2,0) \quad 3 x^{2}+4 y^{2}=48\)

Step-by-Step Solution

Answer

The equation of the ellipse is \(\frac{x^2}{16} + \frac{y^2}{12} = 1\).

Step 1: Understand the ellipse equation in standard form

The standard form of an ellipse centered at the origin is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a\) is the semi-major axis and \(b\) is the semi-minor axis. The vertices are at \((\pm a, 0)\) and the foci at \((\pm c, 0)\), where \(c^2 = a^2 - b^2\).

Step 2: Identify semi-major axis, vertices, and foci

Given vertices at \((\pm 4,0)\), the distance from the center to a vertex (\(a\)) is 4. Given foci at \((\pm 2,0)\), the distance from the center to a focus (\(c\)) is 2.

Step 3: Use foci condition to find \(b^2\)

Use the equation \(c^2 = a^2 - b^2\). Substitute \(c = 2\) and \(a = 4\): \((2)^2 = (4)^2 - b^2\). Solve for \(b^2\): \(4 = 16 - b^2\), therefore \(b^2 = 12\).

Step 4: Formulate the equation of the ellipse

Plug in values for \(a^2\) and \(b^2\) into the standard ellipse equation: \(\frac{x^2}{4^2} + \frac{y^2}{b^2} = 1\). This results in \(\frac{x^2}{16} + \frac{y^2}{12} = 1\).

Step 5: Verify the proposed alternate equation

The alternate equation provided is \(3x^2 + 4y^2 = 48\). To compare, express it in standard form: divide the entire equation by 48: \(\frac{3x^2}{48} + \frac{4y^2}{48} = 1\). Simplifying gives \(\frac{x^2}{16} + \frac{y^2}{12} = 1\), which matches the standard ellipse equation we found.

Key Concepts

Ellipse Properties

Ellipses are fascinating geometric shapes that look like stretched circles. They have two main axes: the major and minor axes. The ellipse has unique properties that distinguish it from circles and other conic sections.

Some key properties of an ellipse include:

• The sum of the distances from any point on the ellipse to the two foci is constant.
• The longest diameter of the ellipse is called the major axis, while the shortest is the minor axis.
• Ellipses have two symmetry axes: one along the major axis and the other along the minor axis.

An ellipse centered at the origin has its major axis aligned with the x-axis or the y-axis. This symmetry simplifies working with their equations and geometric properties.

Standard Form of an Ellipse

The standard form of an ellipse's equation helps us understand its shape, size, and orientation. When centered at the origin, the equation takes the form:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

where \(a\) and \(b\) are constants that define the ellipse's axes. Specifically:

• \(a\) represents the semi-major axis.
• \(b\) represents the semi-minor axis.

In cases where \(a > b\), the ellipse is stretched along the x-axis, while if \(b > a\), it is stretched along the y-axis.

This form is crucial as it makes it easy to identify the major and minor axes lengths and to graph the ellipse effortlessly.

Semi-Major and Semi-Minor Axes

The semi-major and semi-minor axes are the lengths that set the shape of an ellipse.

These axes are fundamental in describing the ellipse's size.

• The semi-major axis, represented by \(a\), is half of the longest diameter of the ellipse. This axis runs through the center and extends to the farthest points on the ellipse.
• The semi-minor axis, represented by \(b\), is half of the shortest diameter. It is perpendicular to the semi-major axis.

To calculate these axes in the standard ellipse equation, we use the denominators of the terms:

• If \(a = 4\), then the ellipse's width along the x-axis is dictated by \(4^2 = 16\).
• If \(b = \sqrt{12}\), then the height along the y-axis is determined by \(12\).

This distinction between axes highlights how stretched or compressed an ellipse appears.

Ellipse Vertices and Foci

Vertices and foci are essential points that characterize an ellipse and help in constructing it.

Vertices:

Vertices are the points where the ellipse is the widest. These are located along the major axis:

• For the ellipse in standard form, the vertices lie at \( (rac{x^2}{a^2}), 0) \) or \( (0, rac{y^2}{b^2}) \) depending on its orientation.
• In our example, the vertices are at \( \pm 4\) along the x-axis.

Foci:

The foci (singular: focus) are two fixed points inside the ellipse. The constant sum of distances from any point on the ellipse to the foci is a defining property.

• To locate the foci, use the formula \( c^2 = a^2 - b^2 \).
• In the given problem, \( c \) equals 2, so the foci are found at \( (rac{c^2}{a^2}), 0) = (rac{2^2}{4^2})\).

Understanding the positions of vertices and foci is vital for accurately sketching the ellipse and appreciating its geometric nuances.