   Chapter 12.4, Problem 49E ### Algebra and Trigonometry (MindTap ...

4th Edition
James Stewart + 2 others
ISBN: 9781305071742

#### Solutions

Chapter
Section ### Algebra and Trigonometry (MindTap ...

4th Edition
James Stewart + 2 others
ISBN: 9781305071742
Textbook Problem

# SKILLS47-58 ■ Graphing Shifted Conics Complete the square to deter-mine whether the graph of the equation is an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, find the center, foci, vertices, and lengths of the major and minor axes. If it is a parabola, find the vertex, focus, and directrix. If it is a hyper-bola, find the center, foci, vertices, and asymptotes. Then sketch the graph of the equation. If the equation has no graph, explain why. x 2 − 5 y 2 − 2 x + 20 y = 44

To determine

The graph of the equation and its characteristics.

Explanation

Given:

The given equation is,

x25y22x+20y=44

Approach:

The basic equation of the ellipse is,

(xh)2a2(yk)2b2=1

The center of hyperbola is given by,

(h,k)

The focal length is given by,

c2=b2+a2

The equation of asymptotes is given by,

yk=±ba(xh)

Coordinates of vertex is given by,

(h±a,k)

Calculation:

Consider the given equation,

x25y22x+20y=44

Now complete the square as shown below,

x25y22x+20y=44(x22x)+(5y2+20y)=44(x22x)5(y24y)=44(x22x+1)5(y24y+4)=44+120

Simplify further,

(x1)25(y2)2=25(x1)2255(y2)25=1

Compare the above equation with (xh)2a2(yk)2b2=1

This is the hyperbola with center (1,2).

And,

a2=25a=±5b2=5b=±5

Find the focal length of the ellipse,

Substitute 25 for a2, and 5 for b2 in focal length formula

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