Problem 12 - Precalculus: Mathematics for Calculus

Question

An equation of an ellipse is given. (a) Find the center, vertices, and foci of the ellipse. (b) Determine the lengths of the major and minor axes. (c) Sketch a graph of the ellipse. $$9 x^{2}-54 x+y^{2}+2 y+46=0$$

Step-by-Step Solution

Answer

The center is \((3, -1)\); vertices at \((3, 5), (3, -7)\); foci at \((3, -1\pm4\sqrt{2})\). Major axis is 12, minor axis is 4.

Step 1: Rewrite the Equation in Standard Form

Start by rearranging and grouping the terms: \(9x^2 - 54x + y^2 + 2y = -46\). Complete the square for the \(x\) terms: \(9(x^2 - 6x)\) becomes \(9((x-3)^2 - 9)\), and for the \(y\) terms: \((y+1)^2 - 1\). The equation becomes \(9((x - 3)^2 - 9) + (y + 1)^2 - 1 = -46\). Simplifying gives \(9(x - 3)^2 + (y + 1)^2 = 36\). Divide everything by 36 to standardize: \(\frac{(x - 3)^2}{4} + \frac{(y + 1)^2}{36} = 1\).

Step 2: Identify the Center of the Ellipse

The equation \(\frac{(x - 3)^2}{4} + \frac{(y + 1)^2}{36} = 1\) indicates that the center of the ellipse is located at \((h, k) = (3, -1)\), based on the standard form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\).

Step 3: Determine the Vertices

Since this is a vertical ellipse (because \(b^2 > a^2\)), the vertices are located at \((h, k \pm b)\). Here, \(b = 6\) and \(a = 2\), so the vertices are \((3, -1 + 6) = (3, 5)\) and \((3, -1 - 6) = (3, -7)\).

Step 4: Calculate the Foci

The foci of a vertical ellipse have coordinates \((h, k \pm c)\), where \(c\) is determined by \(c = \sqrt{b^2 - a^2}\). Here, \(b^2 = 36\) and \(a^2 = 4\), so \(c = \sqrt{36 - 4} = \sqrt{32} = 4\sqrt{2}\). The foci are \((3, -1 + 4\sqrt{2})\) and \((3, -1 - 4\sqrt{2})\).

Step 5: Find the Major and Minor Axis Lengths

For an ellipse, the length of the major axis is \(2b\) and the minor axis is \(2a\). Here, \(b = 6\) and \(a = 2\), so the major axis length is \(12\) and the minor axis length is \(4\).

Step 6: Sketch the Ellipse

In the graph, position the center at \((3, -1)\). Locate the vertices at \((3, 5)\) and \((3, -7)\), and mark the endpoints of the minor axis at \((1, -1)\) and \((5, -1)\). The foci at \((3, -1 + 4\sqrt{2})\) and \((3, -1 - 4\sqrt{2})\) are inside the ellipse along the major axis. Draw an ellipse connecting these points.

Key Concepts

Ellipse center

The center of an ellipse plays a crucial role in its geometry. It is the midpoint that balances the elliptical shape symmetrically on both axes. To determine the center from the equation, we need to identify the terms \(x - h\) and \(y - k\) in the standard form of an ellipse, which is \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\).In this equation, the coordinates of the center are \(h, k\). For example, in our specific equation \(\frac{(x - 3)^2}{4} + \frac{(y + 1)^2}{36} = 1\), the terms \(x - 3\) and \(y + 1\) indicate that the ellipse's center is at \( (3, -1) \).Knowing the center is vital as it serves as a reference point for determining other elements like vertices and foci.

Vertices of ellipse

Vertices are the points where the ellipse is widest and longest. They lie on the major axis and are symmetric relative to the center. For an ellipse described by the standard equation \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), if \(b^2 > a^2\), it means the major axis is vertical. Thus, the vertices' coordinates are \( (h, k \pm b) \).In our exercise, \(b = 6\) and \(a = 2\), leading to vertices at \( (3, 5) \) and \( (3, -7) \). These are determined by the center at \( (3, -1) \) and vertically offset by \(b\). These define the ellipse's height.

Foci of ellipse

The foci of an ellipse are two distinct points located along the major axis. They are important as they help maintain a key property of ellipses: the sum of distances from any point on the ellipse to the foci is constant.To find the foci, use the relation \(c = \sqrt{b^2 - a^2}\). Here, since \(b^2 = 36\) and \(a^2 = 4\), we get \(c = \sqrt{32} = 4\sqrt{2}\). These foci are positioned at \( (h, k \pm c)\), which in this example corresponds to \( (3, -1 + 4\sqrt{2}) \) and \( (3, -1 - 4\sqrt{2}) \).This mathematical feature makes the ellipse unique, distinguishing it from other conic shapes.

Ellipse axes lengths

An ellipse has two axes: a major and a minor axis. The major axis is the longest line, extending through the foci, and the minor axis is perpendicular to it. In mathematical terms, for an equation \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), the length of the major axis is \(2b\) if \(b > a\), and the length of the minor axis is \(2a\).In this case, we have \(b = 6\) and \(a = 2\), so:

• The major axis length is \(2 \times 6 = 12\).
• The minor axis length is \(2 \times 2 = 4\).

Understanding the axis lengths helps us visualize the ellipse's dimensions and plot it accurately.

Graphing ellipses

Graphing an ellipse requires understanding of its center, vertices, and foci, and a mental map of its axes. Begin by plotting the center point, which is our anchor. From the center, mark the vertices and endpoints of the major and minor axes:

• Plot the center at \( (3, -1) \).
• Vertices go at \( (3, 5) \) and \( (3, -7) \).
• Endpoints of the minor axis are at \( (1, -1) \) and \( (5, -1) \).

Don't forget to mark the foci at \( (3, -1 + 4\sqrt{2}) \) and \( (3, -1 - 4\sqrt{2}) \) within the ellipse.Connect these points with a smooth, rounded oval shape. Remember, the major axis determines the ellipse's "height," and the minor axis its "width," determining how "squashed" the ellipse appears.