Problem 34 - 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. $$\frac{(x-2)^{2}}{16}+\frac{(y+3)^{2}}{12}=1$$

Step-by-Step Solution

Answer

Answer: The ellipse has the center at (2, -3), lengths of major and minor axes are 8 and \(4\sqrt{3}\), vertices at (6, -3) and (-2, -3), and foci at (4, -3) and (0, -3).

Step 1: Identify the center of the ellipse

The center \((h, k)\) of the ellipse can be found by comparing the given equation with the standard form of an ellipse equation. In our case, $$\frac{(x-2)^2}{16} + \frac{(y+3)^2}{12} = 1$$, we have \(h = 2\) and \(k = -3\). So, the center of the ellipse is \((2, -3)\).

Step 2: Determine the lengths of the axes

The lengths of the major and minor axes are \(2a\) and \(2b\), respectively. From the given equation, we have \(a^2 = 16\) and \(b^2 = 12\). So, \(a = 4\) and \(b = 2\sqrt{3}\). Hence, the lengths of the major and minor axes are \(2a = 8\) and \(2b = 4\sqrt{3}\).

Step 3: Identify the vertices

The vertices are the points on the ellipse that are furthest apart along the major axis. Since the major axis is along the x-axis, we can find the vertices by adding and subtracting the value of \(a\) from the x-coordinate of the center. Accordingly, the vertices are located at \((2+4, -3)\) and \((2-4, -3)\), which are \((6, -3)\) and \((-2, -3)\).

Step 4: Identify the foci

The foci are the points located along the major axis at a distance of \(\sqrt{a^2 - b^2}\) from the center. In this case, we have \(a^2 = 16\) and \(b^2 = 12\), so we get \(c = \sqrt{16-12} = \sqrt{4} = 2\). The foci are located at \((2+2, -3)\) and \((2-2, -3)\), which are \((4, -3)\) and \((0, -3)\).

Step 5: Conclusion

The given equation $$\frac{(x-2)^2}{16} + \frac{(y+3)^2}{12} = 1$$ is an ellipse with center \((2, -3)\), lengths of major and minor axes to be \(8\) and \(4\sqrt{3}\), vertices at \((6, -3)\) and \((-2, -3)\), and foci at \((4, -3)\) and \((0, -3)\).

Key Concepts

Conic Sections

Conic sections are shapes created by slicing a cone with a plane in different ways. These shapes include circles, ellipses, parabolas, and hyperbolas. Each shape has distinct equations and properties that define them. Understanding conic sections is essential in geometry, as they have applications in various fields such as astronomy, physics, and engineering.

In this case, the given equation represents an ellipse. An ellipse is like a stretched circle with two focal points, and it appears when the intersecting plane cuts through a cone at an angle.

Ellipse Equation

The standard form of an ellipse equation is \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\). Here, \((h, k)\) signifies the ellipse's center, while \(a\) and \(b\) represent the semi-major and semi-minor axes. In our equation, \(\frac{(x-2)^2}{16} + \frac{(y+3)^2}{12} = 1\), comparing with the standard form gives \(h = 2\), \(k = -3\), \(a^2 = 16\), and \(b^2 = 12\). Consequently, \(a = 4\) and \(b = 2\sqrt{3}\).

This configuration tells us the ellipse is oriented with its major axis aligned along the x-axis due to the larger value under the \((x-h)^2\) term.

Vertices of Ellipse

Vertices are crucial points on an ellipse along the major axis where the ellipse reaches its maximum width. Calculating these involves the center and the length of the semi-major axis \(a\).

For our ellipse, the vertices are found by adjusting the x-coordinate of the center by \(a\), yielding the coordinates \((6, -3)\) and \((-2, -3)\).

So, these vertices provide vital insight into the size and orientation of the ellipse.

Foci of Ellipse

The foci are two central points on the ellipse from which the sum of the distances to any point on the ellipse remains constant. The foci are inside the ellipse along the major axis and are pivotal in understanding the ellipse's symmetrical properties.

To find the foci, calculate \(c\) using \(c = \sqrt{a^2 - b^2}\). With \(a^2 = 16\) and \(b^2 = 12\), we have \(c = 2\).

This makes the foci positions \((4, -3)\) and \((0, -3)\), providing clear markers of the ellipse's interior structure.

Graphing Ellipses

Graphing an ellipse involves plotting its critical features: the center, vertices, and foci. Begin by marking the center point \((2, -3)\). Then, from this point, measure out \(a\) units to the left and right to locate the vertices, and \(c\) units to place the foci.

Draw the ellipse so it encapsulates these marked features, ensuring it stretches further along the major axis (horizontal in this case).

• Mark the center: \((2, -3)\)
• Plot the vertices: \((6, -3)\), \((-2, -3)\)
• Locate the foci: \((4, -3)\), \((0, -3)\)

These steps help produce an accurate graph representing the beautiful symmetry and shape of an ellipse.