Problem 36 - Contemporary Precalculus

Question

Identify the conic section whose equation is given, and find its graph. If it is a circle, list its center and radius. If it is an ellipse, list its center, vertices, and foci. $$\frac{(x+5)^{2}}{4}+\frac{(y+2)^{2}}{12}=1$$

Step-by-Step Solution

Answer

Question: Determine the center, semi-major axis, semi-minor axis, vertices, and foci of the ellipse represented by the equation $\frac{(x+5)^2}{4}+\frac{(y+2)^2}{12} = 1$. Answer: The ellipse has a center at (-5, -2), a semi-major axis of $2\sqrt{3}$, a semi-minor axis of 2, vertices at (-5+2√3, -2), (-5-2√3,-2), (-5,0), and (-5,-4), and foci at (-5+2√2, -2), and (-5-2√2, -2).

Step 1: Identify the center, semi-major axis, and semi-minor axis of the ellipse

To find the center \((h,k)\) of the ellipse, look at the denominators of the given equation: $$\frac{(x+5)^{2}}{4}+\frac{(y+2)^{2}}{12}=1$$ The center is given by \((-5,-2)\). Since \(12>4\), \(a^{2}=12\) and \(b^{2}=4\). Thus, \(a=\sqrt{12}=2\sqrt{3}\), and \(b=2\).

Step 2: Find vertices of the ellipse

The vertices are given by the points \((h\pm a, k)\) and \((h, k\pm b)\), where \((h,k)\) is the center, \(a\) is the semi-major axis, and \(b\) is the semi-minor axis. The vertices of the ellipse are: $$(-5\pm2\sqrt{3},-2)$$ and $$(-5,-2\pm2)$$ So, the vertices are \((-5+2\sqrt{3},-2)\), \((-5-2\sqrt{3},-2)\), \((-5,-2+2)\), and \((-5, -2-2)\). Simplifying, we have: \((-5+2\sqrt{3},-2)\), \((-5-2\sqrt{3},-2)\), \((-5,0)\), and \((-5,-4)\)

Step 3: Find the foci of the ellipse

The distance from the center to each focus is given by \(c=\sqrt{a^{2}-b^{2}}\). Using \(a=\sqrt{12}\) and \(b=2\) from previous steps, we get: $$ c = \sqrt{(\sqrt{12})^2 - 2^2} = \sqrt{12-4} = \sqrt{8} = 2\sqrt{2} $$ The foci are located at points \((h\pm c,k)\), where \((h,k)\) is the center of the ellipse. Thus, the foci are given by the points \((-5\pm2\sqrt{2}, -2)\). So, the foci are: \((-5+2\sqrt{2},-2)\) and \((-5-2\sqrt{2},-2)\) In conclusion, the given equation represents an ellipse with center \((-5, -2)\), semi-major axis \(2\sqrt{3}\), semi-minor axis \(2\), vertices \((-5+2\sqrt{3},-2)\), \((-5-2\sqrt{3},-2)\), \((-5,0)\), \((-5,-4)\) and foci \((-5+2\sqrt{2},-2)\), \((-5-2\sqrt{2},-2)\).

Key Concepts

Ellipse Equation

When a student first encounters the ellipse equation, it can appear daunting. But with a little breakdown, it becomes much more approachable. An ellipse is a conic section that looks like a stretched out circle. Its standard equation in the Cartesian plane is given by \[ \frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1 \. \]Here, \((h,k)\) is the center of the ellipse, and \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively. If \(a^2 > b^2\), then the ellipse is stretched along the x-axis, and if \(b^2 > a^2\), it is stretched along the y-axis.

Let's consider an example: \[ \frac{(x+5)^{2}}{4} + \frac{(y+2)^{2}}{12} = 1 \. \]The center can be identified as \((-5, -2)\), and by comparing it to the standard equation, we can see that \(a^2 = 12\) and \(b^2 = 4\), which means the ellipse is stretched along the y-axis. Thus, understanding the standard form of an ellipse equation is critical, as it lays out the framework for identifying all other characteristics of the ellipse.

Vertices of an Ellipse

The vertices of an ellipse are crucial points that lie along the major and minor axes, at the farthest distance from the center of the ellipse. For an ellipse with the standard equation \[ \frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1, \. \]the vertices are found at the points \((h \pm a, k)\) and \((h, k \pm b)\), depending on whether the ellipse is oriented horizontally or vertically.

In the example \[ \frac{(x+5)^{2}}{4} + \frac{(y+2)^{2}}{12} = 1, \. \]the vertices are \((-5 \pm \sqrt{12}, -2)\) and \((-5, -2 \pm 2)\), which simplifies to \((-5 \pm 2\sqrt{3}, -2)\) and \((-5, 0)\), \((-5, -4)\). These points help provide the shape and size of the ellipse and are essential in sketching its accurate representation on a coordinate plane.

Foci of an Ellipse

Diving deeper into the anatomy of an ellipse, the foci are another pair of points which play a vital role in the ellipse's definition and properties. These points are located along the major axis, equidistant from the center. The distance from the center to a focus, denoted by \(c\), is calculated using the formula \[ c = \sqrt{a^{2} - b^{2}}, \. \]where \(a\) is the length of the semi-major axis, and \(b\) is the length of the semi-minor axis. Intuitively, the closer the foci are to each other, the more circular the ellipse appears.

In our previous example, we use \(a = \sqrt{12}\) and \(b = 2\) to find that \(c = \sqrt{8} = 2\sqrt{2}\). The foci are then at the points \((-5 \pm 2\sqrt{2}, -2)\). These special points are not just mathematically interesting but are also related to orbital paths in physics, where planets follow elliptical orbits with the sun at one focus. Understanding the foci helps students grasp both the shape of the ellipse and the deeper relationships within conic sections.