Problem 16 - Algebra and Trigonometry

Question

Find the vertices, foci, and eccentricity of the ellipse. Determine the lengths of the major and minor axes, and sketch the graph. $$ 5 x^{2}+6 y^{2}=30 $$

Step-by-Step Solution

Answer

Vertices: \((\pm 2.45, 0)\). Foci: \((\pm 1, 0)\). Eccentricity: \(0.41\).

Step 1: Rewrite the Equation in Standard Form

The given equation is \(5x^2 + 6y^2 = 30\). First, we divide each term by 30 to get the equation in standard form:\[ \frac{x^2}{6} + \frac{y^2}{5} = 1 \]This represents an ellipse centered at the origin \((0, 0)\) with semi-major axis along the y-direction.

Step 2: Identify the Semi-major and Semi-minor Axes

Compare the denominators in the standard form: \( \frac{x^2}{6} + \frac{y^2}{5} = 1 \). Here, 6 is larger than 5, indicating the semi-major axis is along the x-direction.Semi-major axis length \(a = \sqrt{6} \approx 2.45\) and semi-minor axis length \(b = \sqrt{5} \approx 2.24\).

Step 3: Calculate the Eccentricity

For an ellipse, the eccentricity \(e\) is given by the formula:\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]Plugging in the values:\[ e = \sqrt{1 - \frac{5}{6}} = \sqrt{\frac{1}{6}} \approx 0.41 \]

Step 4: Find the Foci

The distance of the foci from the center along the major axis direction is given by:\[ c = ae = \sqrt{6} \cdot \sqrt{\frac{1}{6}} = 1 \]Thus, the coordinates of the foci are \((\pm 1, 0)\).

Step 5: Determine the Vertices

The vertices of the ellipse are located at the ends of the major axis:Vertices along the x-axis, since it's longer: \((\pm \sqrt{6}, 0)\), which is approximately \((\pm 2.45, 0)\).

Key Concepts

Vertices of Ellipse

In an ellipse, the vertices are key points found at the endpoints of the major axis. For our ellipse given by the equation \(5x^2 + 6y^2 = 30\), after conversion to standard form as \(\frac{x^2}{6} + \frac{y^2}{5} = 1\), we identify the semi-major axis lies along the x-direction. This move's identified by comparing the denominators of the standard form, determining which is larger. Hence, the semi-major axis lies in the direction of the largest denominator, 6 in this case.
Vertices mark the farthest points along the major axis, away from the center of the ellipse. For this ellipse, the vertices are located at:

• \((-\sqrt{6}, 0)\), approximately \((-2.45, 0)\)
• \((\sqrt{6}, 0)\), approximately \((2.45, 0)\)

These points reflect how far the ellipse stretches along its longest horizontal direction.

Foci of Ellipse

The foci of an ellipse are two crucial points located inside the ellipse along the major axis. Foci lead to the elliptical shape through their equal sum of distances from any point on the ellipse. For our equation \(5x^2 + 6y^2 = 30\), the foci lie along the x-direction when simplified to standard form. To find them:

• Determine the value \( c \) using \( c = ae \), where \( a \) is the semi-major axis, and \( e \) is the eccentricity.
• For this ellipse, the foci are \((\pm 1, 0)\).

The distance between these foci defines the elliptic shape. Foci always reside within the ellipse, meaning their x-coordinates fall within the range set by the vertices.

Eccentricity of Ellipse

Eccentricity \( e \) is a parameter that measures how much an ellipse deviates from being circular. An ellipse with an eccentricity close to 0 is nearly circular, whereas as \( e \) approaches 1, the ellipse appears more elongated. It's calculated using \\[ e = \sqrt{1 - \frac{b^2}{a^2}} \] \where \( a \) is the semi-major axis, and \( b \) is the semi-minor axis. For our specific ellipse:

• \( e = \sqrt{1 - \frac{5}{6}} = \sqrt{\frac{1}{6}} \approx 0.41 \)

This relatively low value signifies our ellipse maintains a more rounded shape. Eccentricity is a telling feature that defines the ovalness of the elliptical shape, assisting in the clear classification against a circle, which has \( e = 0 \).

Semi-major Axis

The semi-major axis is the longest radius of an ellipse, extending from the center towards a vertex. In mathematical terms, it is usually denoted as \( a \). Given the equation \( \frac{x^2}{6} + \frac{y^2}{5} = 1 \), we know the semi-major axis lies in the x-direction, due to the larger denominator.
The semi-major axis length for our problem is:

• \( a = \sqrt{6} \approx 2.45 \)

Since it dictates the length from the center to a vertex, the semi-major axis entirely affects overall ellipse width across its longest part. This axis length describes half the distance across the ellipse's widest point.

Semi-minor Axis

In contrast, the semi-minor axis is the shortest radius of the ellipse, stretching from the center outward in the direction of the smaller denominator of the standard ellipse equation. For our ellipse, which further simplified to \( \frac{x^2}{6} + \frac{y^2}{5} = 1 \), the semi-minor axis length aligns naturally with the y-direction, observing a smaller denominator in comparison.
The semi-minor axis length is calculated as:

• \( b = \sqrt{5} \approx 2.24 \)

The semi-minor axis delineates the shorter curve from the center to the edge. This axis helps to shape the ellipse, dictating its height and providing a contrast to its length.