Problem 42 - Algebra and Trigonometry Real Mathematics, Real People

Question

(a) find the standard form of the equation of the ellipse, (b) find the center, vertices, foci, and eccentricity of the ellipse, and (c) sketch the ellipse. Use a graphing utility to verify your graph. $$x^{2}+4 y^{2}-6 x+20 y-2=0$$

Step-by-Step Solution

Answer

The standard form of the given ellipse is \((x-3)^2/36 + (y+2.5)^2/22.5 = 1\). The center is at (3,-2.5), the semi-major axis is 6, the semi-minor axis is roughly 4.74, the foci are roughly 3.61 units either side of the center along the major axis and the eccentricity is roughly 0.60.

Step 1: Convert to Standard Form

First, the given equation needs to be arranged into a recognizable standard form for the equation of an ellipse. To do this, group the \(x\) terms and the \(y\) terms together and complete the square for each. The given equation is \(x^{2}+4 y^{2}-6 x+20 y-2=0\). After rearranging, this becomes \((x^2 - 6x) + 4(y^2+5y) = 2\). Completing the square, this translates to \((x-3)^2 + 4(y+2.5)^2 = 2 + 9 + 25\), which simplifies to \((x-3)^2/36 + (y+2.5)^2/22.5 = 1\)

Step 2: Identify key attributes

Now, from the standard form, we may identify the center (h,k), the length of the semi-major and semi-minor axes (a and b), and calculate the foci of the ellipse. In this case, the center (h,k) is at (3,-2.5). The values of a and b are extracted from the denominators of the \(x\) and \(y\) terms, and are \(\sqrt{36} = 6\) and \(\sqrt{22.5} \approx 4.74\) respectively. The foci can be calculated using the formula \(c = \sqrt{a^2 - b^2}\), providing c ≈ 3.61. The eccentricity \(e\) is given by \(e = c/a = 3.61/6 \approx 0.60\).

Step 3: Sketch the Ellipse

Plot the center of the ellipse at (3,-2.5). From the center, move 6 units left/right (span of major axis) and 4.74 units up/down (span of minor axis) to plot the vertices. The foci are situated 3.61 units left and right of the center along the major axis. Sketch the ellipse by drawing an oval shape that intersects the vertices and encloses the foci within. Verifying this sketch with a graphing utility further facilitates understanding.

Step 4: Verify using Graphing Utility

A graphing calculator or online tool can be used to plot the equation of the ellipse to confirm accuracy. The graph should resemble the sketch, with the correct center, vertices, and foci.

Key Concepts

Standard Form of Ellipse Equation

The standard form of an ellipse's equation is essential for understanding its geometry. An ellipse can be represented by the equation \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\). Here, \((h, k)\) represents the center of the ellipse. The terms \(a^2\) and \(b^2\) determine the lengths of the semi-major and semi-minor axes.

• If \(a > b\), the ellipse is wider horizontally.
• If \(b > a\), it's taller vertically.
• \(a\) is always associated with the larger denominator.

Transforming a general equation of an ellipse, like \(x^2 + 4y^2 - 6x + 20y - 2 = 0\), into this standard form often involves completing the square.

Completing the Square

Completing the square is a technique used to rearrange a quadratic equation. It makes finding the ellipse's standard form possible. This process involves making perfect square trinomials.
First, group the \(x\) terms and \(y\) terms separately from the original equation:

• Group: \((x^2 - 6x) + 4(y^2 + 5y) = 2\)
• Complete squares: \((x-3)^2\) for the \(x\) and \(4(y+2.5)^2\) for the \(y\).

Adjust constants on the right side to complete the squares perfectly. The equation becomes \(\frac{(x-3)^2}{36} + \frac{(y+2.5)^2}{22.5} = 1\), giving it a standard form suitable for further analysis.

Ellipse Center and Vertices

Understanding the center and vertices of an ellipse from its standard form helps in visualizing its shape.

• The center \((h, k)\) in this case is \((3, -2.5)\).
• Major axis length is determined by \(2a = \sqrt{36} = 6\) as \(a\).
• Minor axis length is \(2b = \sqrt{22.5} \approx 4.74\) as \(b\).

Plot these on a graph:

• From the center, move \(a\) units along the major axis for horizontal ovals.
• Move \(b\) units along the minor axis for vertical.

Defining these plots helps create an accurate depiction of the ellipse, ensuring smooth transitions between vertices and the center.

Ellipse Eccentricity

Eccentricity is a crucial parameter to describe the oval-like shape of an ellipse. Represented as \(e\), it is calculated using \(e = \frac{c}{a}\), where \(c\) is the distance between the center and each focus.

• To find \(c\), use \(c = \sqrt{a^2 - b^2}\).
• Here, \(c \approx \sqrt{36 - 22.5} = 3.61\).
• Thus, \(e \approx \frac{3.61}{6} \approx 0.60\).

An eccentricity of \(0\) means a perfect circle; an ellipse has \(0 < e < 1\). The closer \(e\) is to 1, the more elongated the ellipse. This parameter informs about the overall shape and tightness of the ellipse.