
(a)
To find:The standard form of the equation of ellipse 16x2+y2=16 .
(a)

Answer to Problem 36E
The standard form of the equation of ellipse 16x2+y2=16 is x212+y242=1 .
Explanation of Solution
Given information:
The equation of the ellipse is 16x2+y2=16 .
Calculation:
Rewrite the given equation of ellipse.
16x2+y2=1616x216+y216=1x21+y216=1x212+y242=1
Therefore, the standard form of the equation of ellipse 16x2+y2=16 is x212+y242=1 .
(b)
To find:The center, vertices, foci and eccentricity of the ellipse 16x2+y2=16 .
(b)

Answer to Problem 36E
The center, vertices, foci and eccentricity of the ellipse 16x2+y2=16 are (0,0) , (0,±4) , (0,±√15) and 0.97 respectively.
Explanation of Solution
Given information:
The equation of the ellipse is 16x2+y2=16 .
Calculation:
As calculated in part(a), the standard form of the equation of ellipse is x212+y242=1 .
Compare the equation of ellipse x212+y242=1 with general equation (x−h)2b2+(y−k)2a2=1 .
This gives a=4 , b=1 , h=0 and k=0 . The major axis of the ellipse is vertical.
Use the formula.
c2=a2−b2c2=16−1c2=15c=√15
Use the formula for eccentricity.
e=ca=√154≈0.97
The center of the ellipse is (h,k)=(0,0) .
The vertices of the ellipse are:
(h,k±a)=(0,0±4)=(0,±4)
The foci of the ellipse are:
(h,k±c)=(0,0±√15)=(0,±√15)
The eccentricity of ellipse is e=0.97 .
Therefore, The center, vertices, foci and eccentricity of the ellipse 16x2+y2=16 are (0,0) , (0,±4) , (0,±√15) and 0.97 respectively.
(c)
To graph:The ellipse of equation 16x2+y2=16 .
(c)

Answer to Problem 36E
The graph of the ellipse 16x2+y2=16 is shown in Figure(1).
Explanation of Solution
Given information:
The equation of the ellipse is 16x2+y2=16 .
Calculation:
The graph of the ellipse 16x2+y2=16 is shown below.
Figure(1)
Verify the result graphically:
The graph of the equation 16x2+y2=16 with the help of graphing utility.
Figure(2)
As both the graphs gives the same values at each point.
Therefore, both the graphs are similar.
Chapter 9 Solutions
Precalculus with Limits: A Graphing Approach
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