Problem 24 - Algebra and Trigonometry

Question

\(23-34\) Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, find the center, foci, vertices, and lengths of the major and minor axes. If it is a parabola, find the vertex, focus, and directrix. If is a hyperbola, find the center, foci, vertices, and asymptotes. Then sketch the graph of the equation. If the equation has no graph, explain why. $$ 9 x^{2}-36 x+4 y^{2}=0 $$

Step-by-Step Solution

Answer

The equation represents an ellipse with center \((2, 0)\), vertices at \((2, 3)\), \((2, -3)\), \((4, 0)\), \((0, 0)\), and foci at \((2, \pm \sqrt{5})\).

Step 1: Group the Terms

First, we rearrange the equation to group the terms involving each variable together.\[9x^2 - 36x + 4y^2 = 0\]

Step 2: Complete the Square for x

For the x terms, factor out the 9:\[9(x^2 - 4x) + 4y^2 = 0\] To complete the square, take half of the coefficient of x (which is -4), square it, yielding 4, and add and subtract it inside the bracket:\[9((x^2 - 4x + 4) - 4) + 4y^2 = 0\] This simplifies to:\[9((x - 2)^2 - 4) + 4y^2 = 0\] \[9(x - 2)^2 - 36 + 4y^2 = 0\]

Step 3: Simplify and Arrange

Add 36 to both sides of the equation to isolate the completed square:\[9(x - 2)^2 + 4y^2 = 36\] Divide the entire equation by 36 to put it in a recognizable conic form:\[\frac{9(x - 2)^2}{36} + \frac{4y^2}{36} = 1\]This simplifies to:\[\frac{(x - 2)^2}{4} + \frac{y^2}{9} = 1\]

Step 4: Identify the Conic

The equation \[\frac{(x - 2)^2}{4} + \frac{y^2}{9} = 1\] is in the standard form of an ellipse, \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), where \(a^2 = 4\) and \(b^2 = 9\). Since \(b^2 > a^2\), the major axis is vertical.

Step 5: Find the Center, Vertices, and Foci

For the given ellipse, the center is at \((h, k) = (2, 0)\). To find the vertices:- Along the vertical major axis (y-direction), they are \((2, 0 \pm 3) = (2, 3)\) and \((2, -3)\).- Along the minor axis (x-direction), they are \((2 \pm 2, 0) = (4, 0)\) and \((0, 0)\).For the foci, use \(c = \sqrt{b^2 - a^2} = \sqrt{9 - 4} = \sqrt{5}\).- The foci are at \((2, 0 \pm \sqrt{5})\).

Step 6: Identify Lengths of Major and Minor Axes

The major axis of the ellipse, which is vertical, has a length of \(2b = 6\). The minor axis, which is horizontal, has a length of \(2a = 4\).

Key Concepts

Completing the Square

Completing the square is a technique used to transform quadratic equations into a form that makes the properties of conics like ellipses, parabolas, and hyperbolas easier to identify. It involves rearranging the equation so that it includes a perfect square trinomial. To complete the square for the expression involving a particular variable, follow these steps:

• Identify the quadratic term (such as \(x^2\) or \(y^2\)) and the linear term (such as \(ax\) or \(ay\)).
• Factor out the coefficient of the quadratic term if it is not 1. For example, in the expression \(9x^2 - 36x\), factor out a 9 to get \(9(x^2 - 4x)\).
• Take half of the coefficient of the linear term, square it, and add it inside the parentheses to form a perfect square trinomial. In our example, half of -4 (the coefficient of \(x\)) is -2, and squaring -2 gives 4.
• Add and subtract this squared value within the brackets. This ensures the equation remains balanced.

By completing the square, you convert \(x^2 - 4x\) into \((x - 2)^2 - 4\). This step is essential for rewriting the equation into a standard conic form and identifying its properties.

Ellipse Identification

The equation of a conic section can be identified as an ellipse when it takes the form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\). By completing the square and rewriting the equation, you can reveal its conic nature.In the given exercise, the transformed equation \(\frac{(x - 2)^2}{4} + \frac{y^2}{9} = 1\) matches with the standard form of an ellipse. Here, the center of the ellipse is located at \((h, k) = (2, 0)\). Notice that the denominators \(a^2 = 4\) and \(b^2 = 9\) determine the lengths of the axes:

• The term associated with the larger denominator (in this case, 9 which is \(b^2\)) indicates that the major axis is in the direction of the associated variable which is \(y\).
• Since \(b^2 > a^2\), it confirms that the ellipse is stretched vertically along the \(y\)-axis, making it a vertical ellipse.

Recognizing this form helps in further analysis of the ellipse, as it dictates how to calculate its vertices and foci.

Major and Minor Axes

Ellipses have two main axes: the major and minor axes. These axes are perpendicular to each other and intersect at the center of the ellipse.When determining the lengths of these axes for an ellipse given by the equation \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\):

• The major axis is the longer of the two axes. Its length is given by \(2b\) if the ellipse is vertical (\(b^2 > a^2\)), or \(2a\) if it is horizontal (\(a^2 > b^2\)).
• The minor axis is the shorter axis and is perpendicular to the major axis. Its length is \(2a\) if the ellipse is vertical, or \(2b\) if it is horizontal.

For this ellipse, since it is vertical, the major axis length is \(2 \times 3 = 6\), and the minor axis length is \(2 \times 2 = 4\).Knowing these axes lengths helps visualize the ellipse's shape and guides us in plotting it correctly.

Vertices and Foci

Vertices and foci are key features of an ellipse. They help define its shape and orientation. For an ellipse, vertices are the points where the ellipse intersects its major and minor axes:

• In our exercise, the vertices on the major axis (vertical, y-direction) are located at \((2, 3)\) and \((2, -3)\).
• The vertices on the minor axis (horizontal, x-direction) are \((4, 0)\) and \((0, 0)\).

Foci lie along the major axis and are located further in from the vertices toward the center. They are used to mathematically define an ellipse's shape:

• To find the foci, calculate \(c = \sqrt{b^2 - a^2}\). For this ellipse, \(c = \sqrt{9 - 4} = \sqrt{5}\).
• The foci are then at \((2, 0 + \sqrt{5})\) and \((2, 0 - \sqrt{5})\), positioned symmetrically about the center.

Understanding these points helps in sketching and analyzing the ellipse.