   Chapter 12.4, Problem 53E ### 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

# 47-58 Graphing Shifted Conics Complete the square to determine 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 the lengths of the major and minor axes. If it is parabola, find the vertex, focus, and directrix. If it is hyperbola, find the center, foci, vertices, and asymptotes. Then sketch the graph of the equation. if the equation has no graph, explain why. 16 x 2 − 9 y 2 − 96 x + 288 = 0 .

To determine

The graph of the equation 16x29y296x+288=0 and its characteristics.

Explanation

Given:

The given equation is,

16x29y296x+288=0

Approach:

The basic equation for an hyperbola with a vertical transverse axis which is used is as follows,

y2a2x2b2=1

Where a and b are constant values.

Expression to find vertices is,

V=(0,±a)

Expression to find Foci is,

F=(0,±c),

Where, c2=a2b2

Expression to find Asymptotes is,

y=±ab(xb)

Calculation:

Consider the given equation,

16x29y296x+288=0

Now complete the square as shown below,

16x296x+1441449y2=288[(4x)22(4x)(12)+(12)2]9y2=288+144(4x12)29y2=14412(x3)29y2=144

Further solve the above equation.

16(x3)2144+9y2144=1(x3)29+y216=1y216(x3)29=1

This is an equation of a hyperbola with center (3,0) and a vertical transverse axis.

Its graph will have the same shape as of un-shifted hyperbola.

y216x9=1

Hyperbola with center at origin.

Here,

a2=16

And,

b2=9

Then,

c2=a2+b2c2=16+9c2=25c=±5

Thus the un-shifted ellipse foci are,

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