Problem 11 - Precalculus: Mathematics for Calculus

Question

An equation of an ellipse is given. (a) Find the center, vertices, and foci of the ellipse. (b) Determine the lengths of the major and minor axes. (c) Sketch a graph of the ellipse. $$4 x^{2}+25 y^{2}-50 y=75$$

Step-by-Step Solution

Answer

The center is (0, 1). Vertices: (5, 1), (-5, 1). Foci: (√21, 1), (-√21, 1). Major axis: 10, Minor axis: 4.

Step 1: Rewrite the Equation in Standard Form

Given the equation: \(4x^2 + 25y^2 - 50y = 75\). First, we'll rewrite it so it takes the standard form of an ellipse equation, \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\). Start by isolating the terms with \(y\): \[4x^2 + 25(y^2 - 2y) = 75\].

Step 2: Complete the Square

Complete the square for the \(y\)-terms. Take half of the coefficient of \(y\), square it, and add it inside the parentheses: \[y^2 - 2y\] becomes \((y-1)^2 - 1\). Incorporate this change: \[4x^2 + 25((y-1)^2 - 1) = 75\]. Simplify to: \[4x^2 + 25(y-1)^2 = 100\].

Step 3: Divide by 100 to Match Standard Form

Divide the entire equation by 100 to set it equal to 1: \[\frac{4x^2}{100} + \frac{25(y-1)^2}{100} = 1\]. Simplify the fractions: \[\frac{x^2}{25} + \frac{(y-1)^2}{4} = 1\]. Now the equation is in the form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\).

Step 4: Identify the Center and Vertices

The center \((h, k)\) is at \((0, 1)\). The vertices are determined by adding and subtracting \(a=5\) to the \(x\)-component and \(b=2\) to the \(y\)-component of the center. Thus, the vertices are at \((5, 1), (-5, 1)\).

Step 5: Calculate the Foci

To find the foci, use \(c = \sqrt{a^2 - b^2}\), where \(a = 5\) and \(b = 2\). This gives \(c = \sqrt{5^2 - 2^2} = \sqrt{21}\). The foci are \((\sqrt{21}, 1), (-\sqrt{21}, 1)\).

Step 6: Determine the Lengths of the Major and Minor Axes

The major axis is 2 times the length of \(a\) (horizontal axis): Major axis length = 2 \(\times 5 = 10\). The minor axis is 2 times the length of \(b\): Minor axis length = 2 \(\times 2 = 4\).

Step 7: Sketch the Graph

Sketch the ellipse with the center at \((0, 1)\). Since the major axis is horizontal, plot the vertices at \((5, 1)\) and \((-5, 1)\). The minor axis is vertical, so plot points \((0, 3)\) and \((0, -1)\). Add the foci at \((\sqrt{21}, 1)\) and \((-\sqrt{21}, 1)\) as reference points.

Key Concepts

Center of Ellipse

The center of an ellipse is a crucial point that can be seen as the 'middle' of the shape. It serves as the anchor point from where various measurements and characteristics of the ellipse are determined. For an ellipse represented by the equation \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), the center is located at the point \((h, k)\). In the given problem, we identify the center of the ellipse by comparing its equation \(\frac{x^2}{25} + \frac{(y-1)^2}{4} = 1\) to the standard form. Here, we can directly see that \(h = 0\) and \(k = 1\), so the center of the ellipse is at \((0, 1)\). Understanding the center helps us place the ellipse correctly when graphing it.

Vertices of Ellipse

The vertices of an ellipse are the points where the ellipse intersects its major axis. They represent the farthest and closest points from the center along the major axis. From the standard form equation, the length of the semi-major axis is represented by \(a\), and in this problem, \(a = 5\). Therefore, the vertices can be found by moving \(a\) units along the major axis direction (horizontal in this case), from the center. Thus, the vertices are located at \((5, 1)\) and \((-5, 1)\). These points are significant when you graph the ellipse because they help define its overall size and orientation.

Foci of Ellipse

The foci are two specific points inside the ellipse, and one of the unique properties of an ellipse is that the sum of the distances from any point on the ellipse to the foci is constant. To find the foci, we calculate \(c = \sqrt{a^2 - b^2}\), where \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively. In this exercise, \(a = 5\) and \(b = 2\). Calculating, we find that \(c = \sqrt{21}\). The foci are positioned symmetrically along the major axis from the center, at \((\sqrt{21}, 1)\) and \((-\sqrt{21}, 1)\). Including the foci in your graph is essential since they help verify the ellipse's properties geometrically.

Major and Minor Axes

The major and minor axes define the longest and shortest diameters of an ellipse. These axes are perpendicular to each other, with the major axis always being longer than the minor axis. For an equation \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), the lengths of these axes can be found as follows:

• Major axis length: \(2a\), which in our case is \(2 \times 5 = 10\).
• Minor axis length: \(2b\), which here is \(2 \times 2 = 4\).

Knowing these lengths is key for accurately sketching the ellipse, since they define its width and height on a graph.

Graphing Ellipses

Graphing an ellipse involves correctly positioning its center, vertices, and foci, as well as drawing its shape following the major and minor axes. Start by plotting the center at \((0, 1)\). Next, mark the vertices \((5, 1)\) and \((-5, 1)\), which lie onthe horizontal major axis. For a clearer depiction, also locate the minor axis endpoints at \((0, 3)\) and \((0, -1)\), derived from moving \(b\) units up and down from the center. Once these points are marked, sketch the smooth, oval-shaped curve that forms the ellipse, ensuring it passes through the vertices and remains symmetric about both axes. Including the calculated foci \((\sqrt{21}, 1)\) and \((-\sqrt{21}, 1)\) can serve as additional guidance to verify the ellipse's dimensions and orientation.