Determine the center, vertices, foci and asymptotres of the hyperbola x249−y232=1. Then, sketch its graph
The hyperbola has the form x2a2−y2b2=1 with center at origin and horizontal transverse axis since the denominator
of x2 is positive. This gives a2=49 and b2=32, so a=7,b=4√2 and c=√a2+b2=√49+32=9
Then, the following are determined as
center (h,k)→(0,0)vertices (±a,0)→(±7,0)foci (±c,0)→(±9,0)asymptote y=±bax→y=±4√27x
Therefore, the graph is
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