Problem 35 - 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. $$\frac{(x+1)^{2}}{16}+\frac{(y-4)^{2}}{8}=1$$

Step-by-Step Solution

Answer

Question: Identify the type of conic section represented by the given equation and provide specific information about the conic: Equation: $$\frac{(x+1)^{2}}{16}+\frac{(y-4)^{2}}{8}=1$$ Answer: The given equation represents an ellipse. The ellipse has a center at (-1, 4), vertices at (3, 4) and (-5, 4), and foci at \((-1 + 2\sqrt{2}, 4)\) and \((-1 - 2\sqrt{2}, 4)\).

Step 1: Identify the type of conic section

To determine the type of conic section the equation represents, we look at its general form. We have a general equation $$\frac{(x+1)^{2}}{16}+\frac{(y-4)^{2}}{8}=1$$ The equation has both x and y squared terms added together, and both terms have different denominators. This equation represents an ellipse.

Step 2: Find the center of the ellipse

The equation of an ellipse in standard form is: $$\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$$ Where (h, k) is the center of the ellipse. Comparing the given equation with the standard form, we can see that h = -1 and k = 4. So, the center of the ellipse is (-1, 4).

Step 3: Determine the values of a and b

From the given equation, we have: $$\frac{(x+1)^2}{16} + \frac{(y-4)^2}{8} = 1$$ Comparing it to the standard form, we can find that \(a^2 = 16\) and \(b^2 = 8\). So, \(a = 4\) and \(b = 2\sqrt{2}\).

Step 4: Find the vertices of the ellipse

The vertices of the ellipse lie along the major axis (which is along the x-axis in this case) and are at a distance 'a' from the center. Since the center is at (-1, 4) and \(a = 4\), the vertices of the ellipse are: \((-1 + 4, 4) = (3, 4)\) and \((-1 - 4, 4) = (-5, 4)\)

Step 5: Find the foci of the ellipse

To find the foci of the ellipse, we need to first find the distance "c" between the center and the foci. The relationship between a, b, and c is given by: $$c^2 = a^2 - b^2$$ So, \(c^2 = 16 - 8 = 8 \Rightarrow c = 2\sqrt{2}\). The foci are also on the major axis, thus their x-coordinate is at a distance 'c' from the center. So, the foci of the ellipse are: \((-1 + 2\sqrt{2}, 4)\) and \((-1 - 2\sqrt{2}, 4)\). Now we have all the necessary information about the ellipse: 1. The conic section is an ellipse. 2. Center: (-1, 4) 3. Vertices: (3, 4) and (-5, 4) 4. Foci: \((-1 + 2\sqrt{2}, 4)\) and \((-1 - 2\sqrt{2}, 4)\)

Key Concepts

Conic Sections

Conic sections are curves obtained by intersecting a plane with a cone. Depending on the angle and position of the intersection, we can get different shapes. These shapes include circles, ellipses, parabolas, and hyperbolas. Each type of conic has its own distinct equation and properties.

• A **circle** is formed when the plane cuts perpendicular to the cone's axis.
• An **ellipse** is formed when the plane cuts through the cone at an angle, but not steep enough to intersect the opposite side.
• A **parabola** results when the plane is parallel to the side of the cone.
• A **hyperbola** is formed when the plane cuts through both halves of the cone.

To identify the type of conic section represented by an equation, one must examine the squared terms and their coefficients.

Ellipse Properties

An ellipse is a type of conic section that looks like an elongated circle. The key properties that define an ellipse involve its axes, center, vertices, and foci. These properties determine the shape and orientation of the ellipse.

• The **major axis** is the longest diameter of the ellipse, while the **minor axis** is the shortest.
• The center of the ellipse is the midpoint of both axes.
• The **vertices** are the points on the ellipse that lie along the major axis.
• The **foci** are two points inside the ellipse such that the sum of the distances from any point on the ellipse to the foci is constant.

Understanding these characteristics helps in drawing and analyzing the ellipse accurately.

Standard Form of Ellipse

The standard form of an ellipse's equation makes it easy to identify its main characteristics, such as its center, axes, vertices, and foci. The equation of an ellipse in standard form is:\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]Here,

• (h, k) is the center of the ellipse.
• \(a^2\) and \(b^2\) are squared terms that represent the lengths of the major and minor axes. If \(a > b\), the major axis is horizontal. If \(b > a\), the major axis is vertical.

By comparing any given ellipse equation to this standard form, you can identify these properties and better understand the figure.

Vertices

The vertices of an ellipse are important as they define the endpoints of the major axis, giving insight into the size and stretch of the ellipse. To find the vertices:

• Determine the center of the ellipse from its equation in standard form.
• Calculate the distance 'a' from the center to the vertices along the major axis.
• For a horizontal major axis with center (h, k), the vertices are at (h ± a, k).
• For a vertical major axis, the vertices are at (h, k ± a).

In the example provided, the vertices are calculated considering that the primary axis is horizontal, thus giving vertices at (3, 4) and (-5, 4), relative to the center (-1, 4).

Foci

The foci of an ellipse are two critical points located along the major axis inside the ellipse itself. The sum of the distances from any point on the ellipse to the two foci is constant, a defining feature of ellipses.Finding the foci involves these steps:

• Calculate the distance 'c' using the formula \(c^2 = a^2 - b^2\), where 'a' and 'b' are the semi-major and semi-minor axes, respectively.
• For a horizontally oriented ellipse, the foci are located at (h ± c, k).
• For vertically oriented ellipses, the foci are at (h, k ± c).

In our example, for a horizontal major axis, the foci are located at \((-1 + 2\sqrt{2}, 4)\) and \((-1 - 2\sqrt{2}, 4)\), reflecting their symmetry across the x-axis.