Problem 7 - 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. $$ 9 x^{2}+4 y^{2}=36 $$

Step-by-Step Solution

Answer

Vertices: (0,3), (0,-3); Foci: (0,√5), (0,-√5); Eccentricity: √5/3.

Step 1: Convert to Standard Ellipse Equation Form

We start by converting the given equation to the standard form of an ellipse. An ellipse is in standard form if it looks like \[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \]. Given equation is \( 9x^{2} + 4y^{2} = 36 \). Divide the entire equation by 36 to normalize it:\[ \frac{9x^2}{36} + \frac{4y^2}{36} = 1 \]This simplifies to:\[ \frac{x^2}{4} + \frac{y^2}{9} = 1 \]

Step 2: Identify Center, Semi-Major and Semi-Minor Axes

The standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) indicates that the center of the ellipse is at \((0,0)\). If \(a^2 < b^2\), \(b\) is the semi-major axis, and \(a\) is the semi-minor axis. Here, \(a^2 = 4\) and \(b^2 = 9\), so \(b = 3\) and \(a = 2\). Hence, the semi-major axis length is 3, and the semi-minor axis length is 2.

Step 3: Determine Vertices

Since the ellipse is oriented vertically (because \(b > a\)), the vertices are aligned along the y-axis. The vertices are positioned at \((0,\pm b)\). Therefore, the vertices are at \((0,3)\) and \((0,-3)\).

Step 4: Calculate Foci

The foci are located at \((h, k \pm c)\), where \(c = \sqrt{b^2 - a^2}\). Here, \(c = \sqrt{9 - 4} = \sqrt{5}\). Thus, the foci are at \((0, \sqrt{5})\) and \((0, -\sqrt{5})\).

Step 5: Eccentricity of the Ellipse

Eccentricity \(e\) is given by \( e = \frac{c}{b} \), where \(c\) is \(\sqrt{5}\) and \(b = 3\). Therefore, eccentricity \( e = \frac{\sqrt{5}}{3} \).

Step 6: Sketch the Ellipse

Plot the center of the ellipse at the origin (0,0). Draw the major axis along the y-axis with a length of 6 (from -3 to 3). Draw the minor axis along the x-axis with a length of 4 (from -2 to 2). The ellipse is elonged along the y-axis. Mark the vertices, and foci accordingly to complete the sketch.

Key Concepts

Standard Form of Ellipse

Understanding the standard form of an ellipse is crucial for solving problems involving ellipses. The standard equation of an ellipse centered at the origin is written as:\[\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\]where \((h, k)\) represents the center of the ellipse, and \(a\) and \(b\) represent the lengths of the semi-major and semi-minor axes, respectively. The axes are associated with the orientation of the ellipse.

• If \(a > b\), the ellipse is elongated horizontally along the x-axis.
• If \(b > a\), the ellipse stretches vertically along the y-axis.

In the given problem, the equation \(9x^2 + 4y^2 = 36\) needs to be converted into standard form by dividing every term by 36, resulting in \(\frac{x^2}{4} + \frac{y^2}{9} = 1\). This reveals that \(a = 2\) and \(b = 3\), indicating the ellipse is vertically oriented.

Vertices of Ellipse

The vertices of an ellipse are the points where the ellipse is widest. These are crucial landmarks on the graph of an ellipse and are aligned along the major axis.In the standard equation \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\), if the ellipse is vertically oriented \((b > a)\), the vertices are given by \((h, k \pm b)\). If it is horizontally oriented \((a > b)\), the vertices would be \((h \pm a, k)\).For the problem, since \(b = 3\) and \(b > a\), the ellipse stretches along the y-axis. The center is at \((0,0)\) and the vertices are positioned at \((0, \pm 3)\), specifically at \((0, 3)\) and \((0, -3)\). These points represent the ends of the major axis.

Foci of Ellipse

The foci of an ellipse are internal points that define its shape. They play a key role in the geometric properties of the ellipse, as any point on the ellipse maintains a constant total distance to both foci.In the standard form \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\), the foci locations depend on the value of \(c\), where \(c = \sqrt{b^2 - a^2}\). For a vertically oriented ellipse \((b > a)\), the foci are located at \((h, k \pm c)\). For this problem, since \(b = 3\) and \(a = 2\), we calculate \(c = \sqrt{9 - 4} = \sqrt{5}\).This means the foci for our ellipse are placed at \((0, \pm \sqrt{5})\). Hence, the precise focus points are \((0, \sqrt{5})\) and \((0, -\sqrt{5})\).

Eccentricity of Ellipse

Eccentricity is an important measure in understanding an ellipse's "roundness". It provides insight into how stretched the ellipse is compared to a perfect circle, characterized by an eccentricity of 0.The formula for eccentricity \(e\) of an ellipse is:\[e = \frac{c}{b}\]where \(c\) is the distance from the center to a focus, and \(b\) is the semi-major axis. For our example, we determined \(c = \sqrt{5}\) and \(b = 3\).Thus, the eccentricity is \(e = \frac{\sqrt{5}}{3}\). This fraction shows that the ellipse is moderately elongated, possessing less roundness than a circle, particularly extending along the y-axis.