Problem 51 - College Algebra with Corequisite Support

Question

For the following exercises, use the given information about the graph of each ellipse to determine its equation. Center (-3,4)\(;\) vertex (1,4)\(;\) one focus: \((-3+2 \sqrt{3}, 4)\)

Step-by-Step Solution

Answer

The equation of the ellipse is \(\frac{(x+3)^2}{16} + \frac{(y-4)^2}{4} = 1\).

Step 1: Identify Ellipse Type

The given center - Center: (-3, 4) - Vertex: (1, 4) - Focus: (\(-3+2 \sqrt{3}, 4\)) indicates the ellipse is horizontal since both vertex and foci are horizontally aligned with the same y-coordinate.

Step 2: Calculate the Distance to Vertex (Semi-major Axis)

The distance from the center to the vertex - Center: (-3, 4) - Vertex: (1, 4)Calculated as follows:\[a = |x_2 - x_1| = |1 - (-3)| = 4\] Hence, the semi-major axis length is 4.

Step 3: Calculate the Distance to Focus (Determine 'c')

The distance from the center to the focus - Center: (-3, 4) - Focus: \((-3 + 2\sqrt{3}, 4)\)Can be calculated as:\[c = |-3 + 2\sqrt{3} - (-3)| = 2\sqrt{3}\]This 'c' helps in calculating the semi-minor axis 'b' through the standard equation relationship.

Step 4: Apply the Ellipse Relationship

For an ellipse, the relationship is given by the equation:\[a^2 = b^2 + c^2\]Substituting the known values:\[16 = b^2 + (2\sqrt{3})^2\]Solving for \(b^2\):\[b^2 = 16 - 4 \times 3 = 4\]Thus, \(b = 2\).

Step 5: Formulate the Ellipse Equation

For a horizontal ellipse centered at \((-3, 4)\), the standard form of the equation is:\[\frac{(x+3)^2}{4^2} + \frac{(y-4)^2}{2^2} = 1\]Simplified, the equation becomes:\[\frac{(x+3)^2}{16} + \frac{(y-4)^2}{4} = 1\]

Key Concepts

Semi-major Axis

An ellipse's semi-major axis is the longest radius extending from its center to the ellipse's edge. This axis is crucial in determining an ellipse's shape. In our exercise, this is a horizontal ellipse centered at \((-3, 4)\), with a vertex at \((1, 4)\). The semi-major axis runs along the x-axis, given the y-coordinates are constant. To find this length, calculate the horizontal distance between the center and the vertex.

Here's how:

• Center: \((-3, 4)\)
• Vertex: \((1, 4)\)
• Distance calculation: \(a = |1 - (-3)| = 4\)

This results in a semi-major axis length of 4. This axis is pivotal for forming the ellipse's equation.

Semi-minor Axis

The semi-minor axis, typically shorter, complements the semi-major axis in defining the ellipse's size and shape. The length of the semi-minor axis can be derived using the ellipse equation relationship: \(a^2 = b^2 + c^2\). Given the semi-major axis \(a\) and distance to the focus \(c\), solve for the semi-minor axis \(b\).

Let's break it down:

• Semi-major axis: \(a = 4\)
• Focus calculation, \(c = 2\sqrt{3}\)
• Equation: \(16 = b^2 + (2\sqrt{3})^2\)
• Solve for \(b^2: b^2 = 16 - 12 = 4\)
• Semi-minor axis \(b: b = 2\)

The calculated semi-minor axis tells us that the ellipse is stretched less vertically than horizontally.

Ellipse Center

The center of an ellipse is the midpoint around which the entire shape is symmetrically plotted. For our equation, the center is specified as \((-3, 4)\). This point is a reference and is integral for writing the ellipse equation in its standard form.

Understanding the center helps in graphing, as it:

• Serves as a pivot point for both axes.
• Determines the ellipse's position on a coordinate plane.

In this exercise, both axes span from the center across a set range horizontally and vertically, making \((-3, 4)\) an essential aspect of finding the final equation.

Graphing Ellipses

Graphed ellipses form enclosed, symmetric shapes on a coordinate plane. To achieve this, one should follow accurate plotting based on center, semi-major axis, and semi-minor axis. Our elliptic equation \(\frac{(x+3)^2}{16} + \frac{(y-4)^2}{4} = 1\) is derived from a horizontal alignment, guided by the formula for ellipse equations.

Steps to graph:

• Identify the center at \((-3, 4)\).
• From the center, move 4 units right and left for the semi-major axis.
• Move 2 units up and down for the semi-minor axis.
• Sketch the symmetric, oval shape connecting these bounds.

When graphing, ensure symmetry about both axes, reflecting the uniform shape of an ellipse, and use this setup to understand spatial relationships in coordinate geometry.