How do you write an equation for the hyperbola with asymptote #y=+-(4x)/3#, contains #(3sqrt(2,)4)#?

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Answer 1 · 2018-01-25T13:44:15

#x^2/9-y^2/16=1#.

Recall that, for the Hyperbola # S : x^2/a^2-y^2/b^2=1#, the

asymptotes are given by, #y=+-b/a*x;" where, "a,b gt 0#.

In our Case, the asymptotes are, #y=+-4/3*x#.

#:. b/a=4/3...........................................................................(1)#.

Next, #(3sqrt2,4) in S rArr (3sqrt2)^2/a^2-4^2/b^2=1#.

#:. 18/a^2-16/b^2=1#.

Multiplying by #b^2#, we get, #18*b^2/a^2-16=b^2#.

By #(1)#, then, #18*(4/3)^2-16=b^2, i.e., b^2=16#.

#:. b=4 (because, b >0), and, (1) rArr a=3, or, a^2=9#.

With #a^2=9, and b^2=16#, the reqd. eqn. of the Hyperbola is,

#x^2/9-y^2/16=1#.