Problem 13 - 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. $$\frac{x^{2}}{25}+\frac{y^{2}}{4}=1$$

Step-by-Step Solution

Answer

Answer: The center of the ellipse is at the origin (0, 0), the vertices are at the points (±5, 0) and (0, ±2), and the foci are at the points (±√21, 0).

Step 1: Identify the conic section

To identify the conic section, we need to look at the equation form: $$\frac{x^{2}}{25}+\frac{y^{2}}{4}=1$$ The given equation is a sum of squares, where every term is non-negative. This is the general form of an ellipse: $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

Step 2: Find the center

The center of an ellipse with equation \(\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\) is \((h,k)\). Since the given equation is \(\frac{x^2}{25}+\frac{y^2}{4}=1\), we have \(h=0\) and \(k=0\). Therefore, the center is at the origin \((0,0)\).

Step 3: Find the vertices

For an ellipse with equation \(\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\), the vertices are \((h\pm a,k)\) and \((h,k\pm b)\). Our ellipse has \(a^2 = 25\) and \(b^2 = 4\), and its center is \((0,0)\). So, \(a = 5\) and \(b = 2\). The vertices are: \((\pm5, 0)\) and \((0, \pm2)\).

Step 4: Find the foci

To find the foci of the ellipse, we need the lengths of the major axis and the minor axis. Our major axis has a length of \(2a = 2\cdot5 = 10\), and the minor axis has a length of \(2b = 2\cdot2 = 4\). Now we find the distance between the center and each focus, which is given by \(c\). We use the ellipse equation: \(c^2 = a^2 - b^2\). Plugging in our values, we get: $$c^2 = 25 - 4 = 21$$ So, \(c = \sqrt{21}\). The foci of the ellipse are \((\pm\sqrt{21},0)\).

Step 5: Graph the ellipse

To graph the ellipse: 1. Plot the center at \((0,0)\) 2. Plot the vertices at the points \((\pm5,0)\) and \((0,\pm2)\). 3. Plot the foci at the points \((\pm\sqrt{21},0)\). 4. Draw a smooth curve connecting the vertices, passing through the center. The curve should be symmetric about both the x-axis and the y-axis, and the foci should be inside the ellipse. The resulting graph is the required ellipse.

Key Concepts

Ellipse

An ellipse is a type of conic section that looks like an elongated circle. It's defined as the set of all points where the sum of the distances to two fixed points (called the foci) is constant. Ellipses are very common and can be seen in various real-world applications such as planetary orbits. The general equation for an ellipse is \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), where \(a\) is the semi-major axis and \(b\) is the semi-minor axis. This equation shows how the ellipse is stretched along the x and y axes.

• If \(a > b\), the ellipse is wider horizontally.
• If \(b > a\), the ellipse is taller vertically.

Center of Ellipse

The center of an ellipse is a crucial aspect as it determines its position in the coordinate plane. For an equation like \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), \(h\) and \(k\) represent the x and y coordinates of the center, respectively.
In our example equation \(\frac{x^2}{25} + \frac{y^2}{4} = 1\), the center is at the origin because \(h = 0\) and \(k = 0\). The center serves as a reference point for both the vertices and foci of the ellipse. This symmetry helps simplify construction and analysis of the ellipse.

Vertices of Ellipse

Vertices are key points on the ellipse. They mark the longest and shortest diameters. For an ellipse, vertices are found by adding and subtracting the lengths of the semi-major and semi-minor axes from the center's coordinates.
Using the formula \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), vertices are at \((h\pm a,k)\) and \((h,k\pm b)\).

• In the example \(\frac{x^2}{25} + \frac{y^2}{4} = 1\), \(a = 5\) (horizontal) and \(b = 2\) (vertical),
• The vertices are located at \((\pm 5, 0)\) on the x-axis and \((0, \pm 2)\) on the y-axis.

The vertices help determine the shape and orientation of the ellipse.

Foci of Ellipse

The foci of an ellipse lie along its major axis. They are located inside the ellipse itself and are crucial for its definition. The distance from the center to each focus is found with the equation \(c^2 = a^2 - b^2\), where \(a\) is the semi-major axis and \(b\) is the semi-minor axis.

• For the ellipse in our example, \(c^2 = 25 - 4 = 21\), resulting in \(c = \sqrt{21}\).
• The foci are at positions \((\pm \sqrt{21}, 0)\) along the x-axis.

Understanding the position of the foci is vital, as each point on the ellipse's curve satisfies the condition that the total distance to both foci is constant. This property is essential in understanding the geometry and applications of ellipses.