Determine the vertices, foci and asymptotes of the hyperbola x2−y2+4=0. Then sketch its graph
We can rewrite the equation as
x2−y2+4=0Subtract 4x2−y2=−4Divide both sides by −4y24−x24=1
Notice that the equation has the form y2a2−x2b2=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=2√2. Thus, we obtain
vertices (0,±a)→(0,±2)
foci (0,±c)→(0,±2√2)
asymptote y=±abx→y=±22x=±x
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