Problem 43 - 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. $$12 x^{2}+20 y^{2}-12 x+40 y-37=0$$

Step-by-Step Solution

Answer

The standard form of the ellipse is \((x - 0.5)^2 / 4 + (y - 1)^2 / 2.25 = 1\). The center is at (0.5, 1), the vertices are at (-1.5, 1), (2.5, 1), (0.5, -0.5) and (0.5, 2.5), and the foci at (-0.82, 1) and (1.82, 1). The eccentricity is 0.66.

Step 1: Obtain Standard Form of Ellipse Equation

Rearrange the given equation of the ellipse in the standard form: Group \(x\) and \(y\) terms separately, then complete the squares. Also, divide the whole equation by the constant term. You should get \((x - 0.5)^2 / 4 + (y - 1)^2 / 2.25 = 1\).

Step 2: Identify Center, Vertices, and Foci

The center of the ellipse is given by (h, k). From the standard equation, we see that h = 0.5 and k = 1. Thus, the center is at (0.5, 1).\n\nThe vertices are at (h ± a, k) and (h, k ± b). From the standard equation, we have a = 2 and b = 1.5. Thus, the vertices are at (0.5 - 2, 1) and (0.5 + 2, 1), that is (-1.5, 1) and (2.5, 1), and (0.5, 1 - 1.5) and (0.5, 1 + 1.5), that is (0.5, -0.5) and (0.5, 2.5).\n\nThe focus points are at (h ± c, k), where \(c=\sqrt{a^2-b^2}\) if \(a>b\) or at (h, k ± c), where \(c=\sqrt{b^2-a^2}\) if \(b>a\). Since a>b, the foci are at (0.5 - c, 1) and (0.5 + c, 1). Computing c yields \(c = \sqrt{4 - 2.25} = 1.32\). Thus, the foci are (-0.82, 1) and (1.82, 1).

Step 3: Identify Eccentricity

Since \(a > b\), the eccentricity \(e\) is found by using the formula \(e = \sqrt{1 - (b^2 / a^2)} = \sqrt{1 - (1.5^2 / 2^2)} = 0.66\).

Step 4: Sketch the Ellipse

Sketch a graph with x-axis and y-axis. Mark the center at (0.5, 1), then draw the major axis from (-1.5, 1) to (2.5, 1), and the minor axis from (0.5, -0.5) to (0.5, 2.5). You should also mark the foci at (-0.82, 1) and (1.82, 1), and indicate the distance from center to a vertex as 2 (major axis length) and center to co-vertices as 1.5 (minor axis length). The sketch should look like an elongated circle with horizontal major axis.

Step 5: Verify the Solution

The last step is to verify the drawn ellipse graph using any graphic software or tool by inputting the standard equation of the ellipse. The graph obtained should match the manually drawn sketch.

Key Concepts

Standard Form of Ellipse

The standard form of an ellipse's equation helps us identify important features like its center, axes lengths, and orientation. This form is written as \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\). Here, \((h, k)\) represents the center of the ellipse.
To convert an ellipse equation to this form, follow these steps:

• Group x and y terms separately.
• Complete the square for each group.
• Divide by the constant to make the right-hand side equal to one.

After simplifying and rearranging, the example equation \(12x^2 + 20y^2 - 12x + 40y - 37 = 0\) becomes \(\frac{(x - 0.5)^2}{4} + \frac{(y - 1)^2}{2.25} = 1\).

Center of Ellipse

The center of an ellipse is the point \((h, k)\) in the standard form equation. This point divides the ellipse into two equal halves. It acts as a balancing point for the geometric shape formed.
For the example equation, the center is \((0.5, 1)\), derived directly from \((x - 0.5)^2\) and \((y - 1)^2\) in the equation. Knowing this, you can plot the position accurately on a coordinate grid to start drawing the ellipse.

Ellipse Vertices

Vertices of an ellipse are the points where the ellipse reaches its maximum width or height. For a horizontal ellipse, these are \((h \pm a, k)\) and for a vertical ellipse, \((h, k \pm b)\).
In this case, since the equation is horizontal, \(a = 2\) and \(b = 1.5\), giving vertices at:

• Horizontal: \((-1.5, 1)\) and \((2.5, 1)\)
• Vertical: \((0.5, -0.5)\) and \((0.5, 2.5)\)

These points help define the boundaries of the ellipse on each axis.

Ellipse Foci

The foci are special points inside the ellipse used to define its shape and orientation. They are along the major axis with each focus at \((h \pm c, k)\) for horizontal ellipses and \((h, k \pm c)\) for vertical ones.
The calculation involves the formula \(c = \sqrt{a^2 - b^2}\). For this example, \(c = \sqrt{4 - 2.25} = 1.32\), leading to foci at \((-0.82, 1)\) and \((1.82, 1)\). Foci play a critical role in the geometric definition of ellipses, emphasizing their unique structure.

Eccentricity of Ellipse

Eccentricity \(e\) measures how much an ellipse deviates from being a circle. It's determined by the formula \(e = \sqrt{1 - \frac{b^2}{a^2}}\). A smaller eccentricity indicates an ellipse closer to a circle.
For the given ellipse, \(a = 2\) and \(b = 1.5\), yielding \(e = 0.66\). This value tells us the ellipse is moderately stretched along its major axis, not too circular nor too linear, offering insight into its shape and balance.