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Chapter 7: Problem 49

Graph each ellipse and give the location of its foci. $$9(x-1)^{2}+4(y+3)^{2}=36$$

Short Answer

Expert verified
The center of the ellipse is at the point (1, -3), semi-major axis 'a' is 3, semi-minor axis 'b' is 2. Foci are located at points (1 + sqrt(5), -3), (1 - sqrt(5), -3).

Step by step solution

01

Standard form of ellipse

Rearrange the given equation \(9(x-1)^{2}+4(y+3)^{2}=36\), to match the standard form of an ellipse. Divide both sides of the equation by 36, to get the standard form as follows:\[ \frac{(x-1)^{2}}{4} + \frac{(y+3)^{2}}{9} = 1 \] So, the center of the ellipse is at point (h, k) = (1, -3), 'a' (length of semi-major axis) is equal to \(\sqrt{9}\) i.e. 3, and 'b' (length of semi-minor axis) is equal to \(\sqrt{4}\) i.e. 2.
02

Find the foci

Now we need to find the foci.Use the formula for foci: \[ c = \sqrt{a^2 - b^2} \] So, \[ c = \sqrt{3^2 - 2^2} \]Which simplifies to: \[ c = \sqrt{5} \] The foci are located at a distance of 'c' from the center along the major axis. Therefore, coordinates of the foci are \[ (1+\sqrt{5}, -3) \] and \[ (1-\sqrt{5}, -3) \] .
03

Graph the ellipse

The graph isn't provided here, but it is expected to include an ellipse that is vertically aligned due to a>b, with center at (1, -3), vertices at (1, 0) and (1, -6), and foci at (1+sqrt(5), -3) and (1-sqrt(5), -3).

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Key Concepts

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

Standard Form of Ellipse
One of the most fundamental steps in working with ellipses is rearranging their equation into the standard form.
The standard form of an ellipse is given as:
  • Horizontal Major Axis: \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \)
  • Vertical Major Axis: \( \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \)
Here, \((h, k)\) represents the center of the ellipse. For the given problem, the equation was \(9(x-1)^2 + 4(y+3)^2 = 36\).
By dividing every term by 36, it transforms into:
  • \(\frac{(x-1)^2}{4} + \frac{(y+3)^2}{9} = 1\)
This indicates a vertically oriented ellipse, as the bigger denominator (\(a^2\)) is under the \(y\) term. Understanding this form helps identify the ellipse's parameters and orientation.
Foci of Ellipse
The foci are critical points inside an ellipse that define its shape. Together with the center, they help determine how elongated the ellipse is.
For an ellipse, the distance from the center to each focus (\(c\)) can be found using the relation:
  • \(c = \sqrt{a^2 - b^2}\)
These foci lie along the major axis. In our exercise, with \(a = 3\) and \(b = 2\), we get:
  • \(c = \sqrt{3^2 - 2^2} = \sqrt{5}\)
Hence, the foci of the ellipse are located at
  • \((1 + \sqrt{5}, -3)\)
  • \((1 - \sqrt{5}, -3)\)
These points are symmetrically placed around the center along the ellipse's vertical major axis.
Semi-Major Axis
The semi-major axis is the longest radius of an ellipse. It stretches from the center to the edge of the ellipse along the longest diameter (major axis).
The semi-major axis's length is represented by the term \(a\) in the standard form equation.
In our given equation:
  • \(\frac{(x-1)^2}{4} + \frac{(y+3)^2}{9} = 1\)
We notice the vertical component \((y-k)^2\) has the larger denominator, indicating the vertical alignment of the major axis, with:
  • \(a = \sqrt{9} = 3\)
This means the semi-major axis extends 3 units vertically from the center, resulting in ellipse vertices located at \((1, 0)\) and \((1, -6)\). Understanding the semi-major axis helps in constructing and visualizing the ellipse's overall shape.
Semi-Minor Axis
In contrast to the semi-major axis, the semi-minor axis is the shortest radius of the ellipse. It runs perpendicular to the major axis from the center.
The length of the semi-minor axis is given by the term \(b\) in the ellipse equation.
For the provided exercise:
  • \(\frac{(x-1)^2}{4} + \frac{(y+3)^2}{9} = 1\)
The x-component has a smaller denominator:
  • \(b = \sqrt{4} = 2\)
Thus, the semi-minor axis extends horizontally 2 units from the center.(1, -3), influencing how wide the ellipse is compared to its vertical stretch. Mastery of the semi-minor axis concept is essential for accurately drawing ellipses and determining their symmetry.

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