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Q18.

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Precalculus Mathematics for Calculus
Found in: Page 797
Precalculus Mathematics for Calculus

Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508

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Short Answer

Graphing Ellipses An equation of an ellipse is given. (a) Find the vertices, foci, and eccentricity of the ellipse.

(b) Determine the lengths of the major and minor axes. (c) Sketch a graph of the ellipse.

4x2+y2=16.

(a).The vertices, foci and eccentricity of the ellipse 4x2+y2=16 are: (0,±4),(±2,0), (0,±25) and 52 respectively.

(b). The length of the major axis is 8 and the length of the minor axis is 4 for the ellipse 4x2+y2=16.

(c). Graph of the ellipse 4x2+y2=16 is as follows:

See the step by step solution

Step by Step Solution

a.Step 1. Given information.

The equation of the ellipse, 4x2+y2=16

Step 2. The vertices, foci, and eccentricity of the ellipse.

Converting the equation to standard form of ellipse x2a2+y2b2=1

We get,

4x216+y216=1x24+y216=1x222+y242=1

Comparing it with the standard form we get,

a=2b=4

Since,a>b , the given ellipse is vertical ellipse and its focus lies on y- axis.

The coordinates of vertices are (0,±b),(±a,0)

Thus we get, coordinates of vertices as, (0,±4),(±2,0)

To find the foci, we will use the formula c2=a2+b2

By substituting the values of a,b.

We get,

c2=22+42c2=20c=±25

The coordinates of foci are

(0,±c)=(0,±25)

The eccentricity of ellipse,

e=254e=52

b.Step 1. Given information.

The equation of the ellipse, 4x2+y2=16

Step 2. The lengths of major and minor axes.

The equation of the given ellipse is 4x2+y2=16

Now converting it into standard form of ellipse x2a2+y2b2=1

We will get,

4x216+y216=1x24+y216=1x222+y242=1

We will now compare it with the standard form,

Thus we get,a=2

Since a<b,

The length of the major axis: =2a

=8

Length of the minor axis:= 2b

= 4

c.Step 1. Given information.

The equation of the given ellipse is x2+4y2=16.

Step 2. We have to solve the given equation for y.

By converting it into the standard form of an ellipse

(x-h)2a2+(y-k)2b2=1

We will get,

4x2+y2=164x216+y216=1x24+y216=1(x0)222+(y0)242=1

On comparing the above equation with the standard equation (x-h)2a2+(y-k)2b2=1 we get

The center(h,k) is(0,0) .

a=4b=2

We will count 4 units above and below the center and 2 units to the left and right of the center.

The points are (0,4),(0,-4),(2,0) and (-2,0).

The graph of the ellipse will be,

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