Problem 27 - 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. $$ 4 x^{2}+25 y^{2}-24 x+250 y+561=0 $$

Step-by-Step Solution

Answer

The equation represents an ellipse with center (3, -5), vertices (8, -5) and (-2, -5), and foci at (3+√21, -5) and (3-√21, -5).

Step 1: Move constant to the right

First, we rewrite the given equation and move the constant term over to the right side. The equation is: \[ 4x^2 + 25y^2 - 24x + 250y = -561 \]

Step 2: Group and prepare terms for completing the square

Group the terms involving \(x\) and \(y\):\[ 4(x^2 - 6x) + 25(y^2 + 10y) = -561 \]

Step 3: Complete the square for x terms

The expression \(x^2 - 6x\) can be completed by adding \((\frac{-6}{2})^2 = 9\). Thus:\[ 4(x^2 - 6x + 9 - 9) \]We include and subtract 9 inside the bracket. This transforms to:\[ 4((x - 3)^2 - 9) = 4(x - 3)^2 - 36 \]

Step 4: Complete the square for y terms

Similarly, complete the square for \(y\) by adding \((\frac{10}{2})^2 = 25\):\[ 25(y^2 + 10y + 25 - 25) \]This transforms to:\[ 25((y + 5)^2 - 25) = 25(y + 5)^2 - 625 \]

Step 5: Simplify and isolate squared terms

Substitute the completed squares back into the equation:\[ 4(x - 3)^2 - 36 + 25(y + 5)^2 - 625 = -561 \]Combine constants:\[ 4(x - 3)^2 + 25(y + 5)^2 = 100 \]

Step 6: Divide by 100 to identify the conic

Divide the entire equation by 100:\[ \frac{4(x - 3)^2}{100} + \frac{25(y + 5)^2}{100} = 1 \]Simplify:\[ \frac{(x - 3)^2}{25} + \frac{(y + 5)^2}{4} = 1 \]

Step 7: Identify conic and its properties

This is the standard form of an ellipse. The center is \((3, -5)\), the lengths of the semi-major and semi-minor axes are 5 and 2, respectively. The major axis is horizontal. The vertices are \((8, -5)\) and \((-2, -5)\). To find the foci, use \(c = \sqrt{25 - 4} = \sqrt{21}\). The foci are \((3+\sqrt{21}, -5)\) and \((3-\sqrt{21}, -5)\).

Step 8: Sketch the ellipse

Sketch an ellipse centered at \((3, -5)\) with a horizontal major axis of length 10 (distance between vertices) and vertical minor axis of length 4. Place foci inside the ellipse, evenly spaced on the major axis.

Key Concepts

Ellipse

An ellipse is a significant shape in conic sections often described as an elongated circle. It's a set of points where the sum of the distances to two fixed points (foci) remains constant. The unique nature of an ellipse is well-captured by its standard equation \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]where - \((h, k)\) is the center of the ellipse.- \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively.

The larger of the two values, \(a\) or \(b\), is associated with the major axis, which is the longest diameter of the ellipse. Depending on the values of \(a\) and \(b\), the ellipse can be elongated horizontally or vertically.

Completing the Square

Completing the square is a mathematical technique used to solve quadratic equations, convert them into a specific standard form, or reveal properties of a conic section. In the context of finding the center and other characteristics of an ellipse, it involves making a quadratic expression into a perfect square trinomial. This form makes it simpler to identify transformations and features of the geometric shape.

Here's the basic process for completing the square:

• Start with a quadratic expression, for example, \(x^2 - 6x\).
• Take half of the coefficient of \(x\), square it, and add and subtract it inside the expression: \((\frac{-6}{2})^2 = 3^2 = 9\).
• The adjusted expression becomes \((x - 3)^2 - 9\).

This technique effectively rewrites the quadratic term in a manner that displays the form of the ellipse and indicates shifts from the origin.

Center of an Ellipse

The center of an ellipse is the midpoint between its foci and represents the point around which the ellipse is symmetrically balanced. Based on the standard equation of an ellipse, \( (h, k) \) denotes its center.

In the problem's context, after completing the square, we arrived at the equation:\[ \frac{(x-3)^2}{25} + \frac{(y+5)^2}{4} = 1 \]Here, the center of the ellipse is identified as \((3, -5)\). This placement of the center ensures the geometric properties of the ellipse hold, enabling accurate plotting on a graph.

Understanding where the center lies is crucial for sketching the ellipse and determining the length of its axes and positioning of other key features like vertices and foci.

Vertices of an Ellipse

The vertices of an ellipse mark the endpoints of the major axis, representing the farthest extents of the shape along one dimension. For an equation in standard form \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]- if \(a > b\), the major axis runs horizontally, and the vertices are at \((h \pm a, k)\).- if \(b > a\), then the major axis is vertical, placing the vertices at \((h, k \pm b)\).

In this exercise, since \(a = 5\) and \(b = 2\), the major axis is horizontal given that 5 is greater than 2. Therefore, the vertices based on the center are found at:

• \((3 + 5, -5) = (8, -5)\)
• \((3 - 5, -5) = (-2, -5)\)

Recognizing the vertices is essential for understanding the ellipse's dimensions and symmetry.