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Concept explainers
In Exercises 1-8, graph each ellipse and locate the foci.
x236+y225=1
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To graph: The equation of ellipse x236+y225=1 and locate the foci.
Explanation of Solution
Given information:
The ellipse equation: x236+y225=1
Graph:
Let us consider the following ellipse equation:
x236+y225=1 (I)
And the general equation of the ellipse is as given below:
x2a2+y2b2=1 (II)
And comparing equations (I) and (II) to obtain:
a2=36 And b2=25
And the equation is the general ellipse’s equation with a2=36 and b2=25.
Since, the denominator of the x2-term is greater than the denominator of the y2-term, therefore, the major axis is horizontal.
Simplifying a2=36 and b2=25 as follows to obtain the value of a and b:
a2=36a2=62a=6
And
b2=25b2=52b=5
And for a standard ellipse equation, the vertices of an ellipse are (–a, 0) and (a, 0), the endpoints of the vertical minor axis are (0, –b) and (0, b).
So, the vertices of the ellipse are (–6, 0) and (6, 0), and the endpoints of the vertical minor axis are (0, –5) and (0, 5).
And the foci F1 and F2 are located at the points (–c, 0) and (c, 0), where c2=a2–b2.
Put the values of a2 and b2 in c2=a2–b2 and find the value of c as follows:
c2=a2–b2c2=36−25c2=11c=√11
The foci F1 and F2 are (−√11, 0) and (√11, 0).
Then, plot the endpoints and the foci and trace them to obtain a smooth curve as shown below:
Interpretation:
Thus, the foci of the provided ellipse equation are (−√11, 0) and (√11, 0), the vertices of the ellipse are (–6, 0) and (6, 0), and the endpoints of the vertical minor axis are (0, –5) and (0, 5).
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Chapter 9 Solutions
Precalculus (6th Edition)
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