Problem 51 - Precalculus

Question

Use a graphing utility to graph the ellipse. Find the center, foci, and vertices. (Recall that it may be necessary to solve the equation for \(y\) and obtain two equations.) $$12 x^{2}+20 y^{2}-12 x+40 y-37=0$$

Step-by-Step Solution

Answer

The center of the ellipse is \((\frac{1}{2}, -1)\), the vertices are \((\frac{1}{2}, -1 \pm \sqrt{5/3})\), and the foci are \((\frac{1}{2}, -1 \pm \sqrt{1/4})\).

Step 1: Rewrite the equation

Rearrange the equation by grouping the \(x\) terms and \(y\) terms: \(12x^2 - 12x + 20y^2 + 40y - 37 = 0\). Then, divide the equation by \(12\) to isolate the \(x^2\) term: \(x^2 - x + \frac{5}{3}y^2 + \frac{10}{3}y - \frac{37}{12} = 0\).

Step 2: Complete the square

Complete the square for the \(x\) terms and \(y\) terms separately: \( (x - \frac{1}{2})^{2} - \frac{1}{4} + \frac{5}{3} \left(y + 1\right)^{2} - \frac{5}{3} - \frac{37}{12} = 0 \). Simplify this equation to get the equation of the ellipse in standard form: \( (x - \frac{1}{2})^{2} + \frac{5}{3} \left(y + 1\right)^{2} = 1 \).

Step 3: Identify the center, vertices and foci

From the standard form of the ellipse, we can see that the center of the ellipse is \((\frac{1}{2}, -1)\). The length of the major axis is \(2\sqrt{5/3}\) and the length of the minor axis is \(2\). As this is a vertically-oriented ellipse, the vertices are \((\frac{1}{2}, -1 \pm \sqrt{5/3})\) and the foci are \((\frac{1}{2}, -1 \pm \sqrt{1/4})\).

Key Concepts

Center of Ellipse

The center of an ellipse is a key point that acts as the midpoint between its foci. To find the center in the given exercise, we rewrite the equation of the ellipse in its standard form. This involves completing the square for both the \(x\) and \(y\) terms. In our particular example, after rewriting, we identify the ellipse equation:
\[(x - \frac{1}{2})^2 + \frac{5}{3}(y + 1)^2 = 1\]From this equation, we can see that the center of the ellipse is

• X-coordinate: \( \frac{1}{2} \), from \(( x - \frac{1}{2})^2 \). This means the ellipse is shifted \( \frac{1}{2} \) units to the right on the x-axis from the origin.
• Y-coordinate: \( -1 \), from \(( y + 1 )^2 \). This means the ellipse is shifted 1 unit down on the y-axis from the origin.

Thus, its center is at coordinates \( (\frac{1}{2}, -1) \). Recognizing the center helps in sketching the ellipse accurately on a graph.

Foci of Ellipse

The foci of an ellipse are two internal points, equidistant from the center, and lie along its major axis. These points are critical in defining the ellipse, as the sum of the distances from any point on the ellipse to the foci is constant.In the given exercise, to determine the foci coordinates, we utilize the formula:
\[ c = \sqrt{a^2 - b^2} \]Here, \(a\) is the semi-major axis (longest radius), and \(b\) is the semi-minor axis. Based on the equation:
\[a^2 = \frac{5}{3},\, b^2 = 1\]The value of \(c\) becomes:
\[\sqrt{\frac{5}{3} - 1} = \sqrt{\frac{2}{3}}\]This value helps us find the foci:

• Since this is a vertical ellipse, the foci are along the y-axis. So, add and subtract \(\sqrt{\frac{2}{3}}\) from the center's \(y\)-coordinate, \(-1\).
• Thus, the foci are at \( (\frac{1}{2}, -1 + \sqrt{\frac{2}{3}}) \) and \( (\frac{1}{2}, -1 - \sqrt{\frac{2}{3}}) \).

The position of the foci aids in understanding the shape and alignment of the ellipse.

Vertices of Ellipse

Vertices are the points where the ellipse is widest and are aligned along its major axis. They can be found directly using the semi-major axis length from the center.For a vertically oriented ellipse, like in this exercise, the vertices are calculated by moving from the center along the y-axis using the semi-major axis length of \(\sqrt{\frac{5}{3}}\).

• The first vertex is found by adding \(\sqrt{\frac{5}{3}}\) to the y-coordinate of the center. For our ellipse, it's at \( (\frac{1}{2}, -1 + \sqrt{\frac{5}{3}}) \).
• The second vertex is determined by subtracting \(\sqrt{\frac{5}{3}}\) from the center's y-coordinate, giving us the point \( (\frac{1}{2}, -1 - \sqrt{\frac{5}{3}}) \).

Locating these vertices helps sketch how far the ellipse stretches along its major axis, completing its graph.