Problem 15 - Precalculus with Calculus Previews

Question

Find the center, foci, vertices, endpoints of the minor axis, and eccentricity of the given ellipse. Graph the ellipse. $$ 5(x-1)^{2}+3(y+2)^{2}=45 $$

Step-by-Step Solution

Answer

Center: (1, -2); Vertices: (1, 1.87) & (1, -5.87); Minor Axis Endpoints: (4, -2) & (-2, -2); Foci: (1, -0.55) & (1, -3.45); Eccentricity: \(\frac{\sqrt{6}}{\sqrt{15}}\).

Step 1: Rewrite the Equation in Standard Form

Start by dividing the entire equation by 45 so that it equals 1. This gives us: \[ \frac{5(x-1)^{2}}{45} + \frac{3(y+2)^{2}}{45} = 1 \]Simplify the fractions:\[ \frac{(x-1)^{2}}{9} + \frac{(y+2)^{2}}{15} = 1 \]This is the standard form of the ellipse equation \( \frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1 \), where \((h, k)\) is the center.

Step 2: Identify the Center

From the standard form \( \frac{(x-1)^{2}}{9} + \frac{(y+2)^{2}}{15} = 1 \), it is evident that the center of the ellipse is \((h, k) = (1, -2)\).

Step 3: Determine Vertices and Orientation

Since \(9 < 15\), the larger denominator is under the \(y\)-term. This means the ellipse is vertically oriented. The lengths of the semi-major and semi-minor axes are \(b = \sqrt{15}\) and \(a = \sqrt{9} = 3\), respectively.Thus, the vertices are at \((1, -2 \pm \sqrt{15})\) which simplifies to approximately \((1, -2 \pm 3.87)\), or \( (1, 1.87) \) and \( (1, -5.87) \).

Step 4: Find Endpoints of the Minor Axis

The endpoints of the minor axis, being perpendicular to the major axis, will occur at the center \(\pm a\) horizontally. Therefore, they are located at \((1 \pm 3, -2)\), which simplifies to \((4, -2)\) and \((-2, -2)\).

Step 5: Calculate Eccentricity and Foci

The eccentricity \(e\) is found using the formula \(e = \frac{c}{b}\), where \(c = \sqrt{b^{2} - a^{2}} = \sqrt{15 - 9} = \sqrt{6}\). Thus, the eccentricity is \(e = \frac{\sqrt{6}}{\sqrt{15}}\).The foci are along the major axis at \((1, -2 \pm \sqrt{6})\), or approximately \((1, -0.55)\) and \((1, -3.45)\).

Step 6: Graph the Ellipse

To graph the ellipse:- Plot the center at \((1, -2)\).- Draw the major axis using the vertices at \((1, 1.87)\) and \((1, -5.87)\).- Draw the minor axis using the endpoints at \((4, -2)\) and \((-2, -2)\).- Mark the foci at approximately \((1, -0.55)\) and \((1, -3.45)\).- Sketch the ellipse through these points, ensuring it is longer vertically.

Key Concepts

Equation of an Ellipse

To fully understand the equation of an ellipse, we start by expressing it in its standard form: \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]This equation describes an ellipse with its center located at \((h, k)\).

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

In our exercise, the original equation\[5(x-1)^2 + 3(y+2)^2 = 45\]was simplified to\[\frac{(x-1)^2}{9} + \frac{(y+2)^2}{15} = 1\]by dividing through by 45. The center \((h, k)\) is found to be \((1, -2)\).This form allows us to quickly identify key characteristics of an ellipse, such as its orientation and size.

Eccentricity of an Ellipse

Eccentricity, denoted as \( e \), is a measure that describes how elongated an ellipse is. It ranges between 0 and 1.

• When \( e = 0 \), the ellipse is a perfect circle.
• The closer \( e \) is to 1, the more elongated the ellipse.

To find the eccentricity, use the formula:\[e = \frac{c}{b}\] where \( c \) is the distance from the center to each focus, and \( b \) is the length of the semi-major axis. In our solution, \( c = \sqrt{6} \), calculated from \( c = \sqrt{b^2 - a^2} = \sqrt{15 - 9} \).Therefore, the eccentricity becomes:\[e = \frac{\sqrt{6}}{\sqrt{15}}\]This formula shows how the shape of the ellipse is slightly stretched compared to a circle.

Graphing Ellipses

Graphing an ellipse involves several steps:

• Plot the center, which is \((1, -2)\) in this exercise.
• Identify and sketch the major and minor axes. The vertices for the major axis were found at \((1, 1.87)\) and \((1, -5.87)\).
• The endpoints of the minor axis, perpendicular to the major, are at \((4, -2)\) and \((-2, -2)\).
• Place the foci along the major axis, approximately at \((1, -0.55)\) and \((1, -3.45)\).

Once these critical points are marked, connect them in a smooth, oval shape, ensuring that the vertical length is more than the horizontal, which accurately reflects the properties of a vertical ellipse. Remember, the symmetry is key in creating an accurate graph.

Ellipse Vertices and Foci

Vertically oriented ellipses have their major axis aligned along the y-axis and therefore, they extend vertically. In such ellipses:

• Vertices mark the endpoints of the major axis. In our problem, they are located at \((1, -2 \pm \sqrt{15})\), or approximately \((1, 1.87)\) and \((1, -5.87)\).
• The foci, also located along the major axis, are crucial for understanding the shape of the ellipse. These points for our ellipse are \((1, -2 \pm \sqrt{6})\) or about \((1, -0.55)\) and \((1, -3.45)\).

The vertices and foci are critical as they dictate both the overall size and shape of the ellipse, emphasizing its elongated direction and providing the precise path for light and sound reflecting properties in practical applications.