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Chapter 8: Problem 51

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: \((-3+2 \sqrt{3}, 4)\)

Short Answer

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

Step by step solution

01

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.
02

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.
03

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.
04

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, \(b = 2\).
05

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:\[\frac{(x+3)^2}{16} + \frac{(y-4)^2}{4} = 1\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

For the following exercises, find a new representation of the given equation after rotating through the given angle. \(4 x^{2}+\sqrt{2} x y+4 y^{2}+y+2=0, \theta=45^{\circ}\)

An object is projected so as to follow a parabolic path given by \(y=-x^{2}+96 x,\) where \(x\) is the horizontal distance traveled in feet and \(y\) is the height. Determine the maximum height the object reaches.

For the following exercises, graph the given conic section. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse, label the vertices and foci. If it is a hyperbola, label the vertices and foci. \(r=\frac{3}{1+2 \cos \theta}\)

Recall from Rotation of Axes that equations of conics with an \(x y\) term have rotated graphs. For the following exercises, express each equation in polar form with \(r\) as a function of \(\theta\). \(x^{2}+x y+y^{2}=4\)

For the following exercises, determine which conic section is represented based on the given equation. \(2 x^{2}-2 y^{2}+4 x-6 y-2=0\)
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