Sunday, January 11, 2015

College Algebra, Chapter 8, 8.3, Section 8.3, Problem 16

Determine the vertices, foci and asymptotes of the hyperbola x2y2+4=0. Then sketch its graph

We can rewrite the equation as


x2y2+4=0Subtract 4x2y2=4Divide both sides by 4y24x24=1


Notice that the equation has the form y2a2x2b2=1. Since the y2-term is positive, then the hyperbola has a vertical transverse axis; its vertices and foci are located on the y-axis, since a2=4 and b2=4, we get a=2 and b=2 and c=a2+b2=22. Thus, we obtain

vertices (0,±a)(0,±2)

foci (0,±c)(0,±22)

asymptote y=±abxy=±22x=±x

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