Problem 35
Question
Find an equation for each ellipse. Center \((5,2) ;\) minor axis vertical, with length \(8 ; c=3\)
Step-by-Step Solution
Verified Answer
The equation of the ellipse is \(\frac{(x-5)^2}{25} + \frac{(y-2)^2}{16} = 1\).
1Step 1: Understanding the Problem
We are given an ellipse with a center at \((5, 2)\), a vertical minor axis with a length of 8, and \(c = 3\). We need to find the equation of this ellipse.
2Step 2: Identifying the Formula for an Ellipse
For an ellipse, the general equation is \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\) where \((h, k)\) is the center. Since the minor axis is vertical, \(a\) is along the x-axis and \(b\) is along the y-axis.
3Step 3: Determine the Semi-minor Axis Length (b)
The length of the minor axis is 8, so the semi-minor axis (\(b\)) is half of 8, which is 4. Thus, \(b = 4\).
4Step 4: Calculate the Semi-major Axis Length (a) Using c
For an ellipse, \(c^2 = a^2 - b^2\). Given that \(c = 3\), we substitute and find: \[ 3^2 = a^2 - 4^2 \] \[ 9 = a^2 - 16 \] \[ a^2 = 25 \] Hence, \(a = 5\).
5Step 5: Write the Equation of the Ellipse
Since the center is \((5, 2)\), \(a = 5\), and \(b = 4\), the equation of the ellipse is: \[ \frac{(x-5)^2}{25} + \frac{(y-2)^2}{16} = 1 \]
Key Concepts
Semi-major axisSemi-minor axisEllipse centerVertical minor axis
Semi-major axis
In the context of an ellipse, the semi-major axis is the longest radius that extends from the center to the edge of the ellipse. This axis represents half of the major axis' full length.
In an ellipse equation of the form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), the semi-major axis is represented by \(a\). When the minor axis is vertical, \(a\) measures along the x-axis, helping to form the longest width of the ellipse.
To determine \(a\), we use the relationship \(c^2 = a^2 - b^2\), where \(c\) is the distance from the center to one of the foci. Given \(c = 3\) and a semi-minor axis \(b = 4\), we initially set up the equation:\
In an ellipse equation of the form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), the semi-major axis is represented by \(a\). When the minor axis is vertical, \(a\) measures along the x-axis, helping to form the longest width of the ellipse.
To determine \(a\), we use the relationship \(c^2 = a^2 - b^2\), where \(c\) is the distance from the center to one of the foci. Given \(c = 3\) and a semi-minor axis \(b = 4\), we initially set up the equation:\
- \(3^2 = a^2 - 4^2\)
- \(9 = a^2 - 16\)
- \(a^2 = 25\)
Semi-minor axis
The semi-minor axis of an ellipse is the shortest radius that extends from the center to the edge. It is half the length of the minor axis.
In the equation \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), the semi-minor axis is represented by \(b\). When the minor axis is vertical, \(b\) measures along the y-axis, defining the ellipse's height.
From the given problem, the minor axis has a total length of 8, so the semi-minor axis \(b\) is:\
In the equation \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), the semi-minor axis is represented by \(b\). When the minor axis is vertical, \(b\) measures along the y-axis, defining the ellipse's height.
From the given problem, the minor axis has a total length of 8, so the semi-minor axis \(b\) is:\
- \(b = \frac{8}{2} = 4\)
Ellipse center
The center of an ellipse is a crucial point indicating its midpoint extending in all directions. It provides the base coordinates for positioning the ellipse in relation to the Cartesian plane.
The general form of an ellipse's equation \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\) uses \((h,k)\) to specify the ellipse's center. Here, \(h\) is the x-coordinate and \(k\) the y-coordinate of the center.
For the given exercise, the ellipse is centered at \((5, 2)\). Thus, we have:
The general form of an ellipse's equation \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\) uses \((h,k)\) to specify the ellipse's center. Here, \(h\) is the x-coordinate and \(k\) the y-coordinate of the center.
For the given exercise, the ellipse is centered at \((5, 2)\). Thus, we have:
- \(h = 5\)
- \(k = 2\)
Vertical minor axis
When we say the minor axis is vertical, it means that the shorter axis of the ellipse runs up and down, parallel to the y-axis.
The orientation of the axes significantly affects the ellipse's equation, as vertical minor axes mean \(b\) (semi-minor axis) affects the y-coordinates. This changes the equation to \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\) where \(a\) is greater than \(b\), aligning \(b\) with the vertical direction.
In our problem, this specificity informs us how to best view and predict the stretch and compression of the ellipse's shape. Understanding that the semi-minor axis is vertical allows for precise determination of the ellipse's proportions and layout, directly impacting our plotting and comprehension.
The orientation of the axes significantly affects the ellipse's equation, as vertical minor axes mean \(b\) (semi-minor axis) affects the y-coordinates. This changes the equation to \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\) where \(a\) is greater than \(b\), aligning \(b\) with the vertical direction.
In our problem, this specificity informs us how to best view and predict the stretch and compression of the ellipse's shape. Understanding that the semi-minor axis is vertical allows for precise determination of the ellipse's proportions and layout, directly impacting our plotting and comprehension.
Other exercises in this chapter
Problem 34
Determine the type of conic section represented by each equation, and graph it, provided a graph exists. $$-4 x^{2}+8 x+y^{2}+6 y=-6$$
View solution Problem 34
Graph each circle by hand if possible. Give the domain and range. $$(x+2)^{2}+(y-5)^{2}=16$$
View solution Problem 35
Graph each circle by hand if possible. Give the domain and range. $$(x-5)^{2}+(y+4)^{2}=49$$
View solution Problem 35
Give two parametric representations for each plane curve. Use your calculator to verify your results. $$y=2 x+3$$
View solution