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Calculus

10th Edition
Ron Larson + 1 other
ISBN: 9781285057095

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Calculus

10th Edition
Ron Larson + 1 other
ISBN: 9781285057095
Textbook Problem

Identifying a Conic In Exercises 7-14, identify the conic, analyze the equation (center, radius, vertices, foci, eccentricity, directrix, and asymptotes, if possible), and sketch its graph. Use a graphing utility to confirm your results.

3 x 2 2 y 2 + 24 x + 12 y + 24 = 0

To determine

The provided conic by analyzing the equation (center, radius, vertices, foci, eccentricity, directrix and asymptotes, if possible), by sketching its graph. To confirm your results, use a graphing utility.

Explanation

Given:

The provided conic equation is 3x22y2+24x+12y+24=0.

Consider the provided equation

3x22y2+24x+12y+24=0

Here, x and y both have squared terms and exactly one of them is negative. Thus, this equation represents a hyperbola.

In order to find the center, write this equation in the form of the standard equation of a hyperbola which is:

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

Rearrange the provided equation:

3x22y2+24x+12y=24

Now, regroup the terms according to the standard form:

3x2+24x2y2+12y=24

Factor out the co-efficient of square terms:

3(x2+8x)2(y26y)=24

Divide the equation by 3:

(x2+8x)23(y26y)=8

Divide the equation by 2:

12(x2+8x)13(y26y)=4

Use the completing squares method:

12(x2+8x+42)13(y26y+32)=4+222323

So,

12(x+4)213(y3)2=4+16293

That is,

(x+4)22(y3)23=1

Rewrite the above equation in standard form as:

(x(4))2(2)2(y3)2(3)2=1

Compare this with the standard equation of a hyperbola with center (h,k)

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

So, h=4 and k=3

Also,

a=2b=3

So, the center of the hyperbola is (4,3)

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