How do you write an equation for the hyperbola with asymptote $y = \pm \frac{4 x}{3}$ $y=+-(4x)/3$ , contains $(3 \sqrt{2,} 4)$ $(3sqrt(2,)4)$ ?

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Answer 1 · 2018-01-25T13:44:15

$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$ $x^2/9-y^2/16=1$ .

Recall that, for the Hyperbola $S : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ $S : x^2/a^2-y^2/b^2=1$ , the

asymptotes are given by, $y = \pm \frac{b}{a} \cdot x; where, a, b > 0$ $y=+-b/a*x;" where, "a,b gt 0$ .

In our Case, the asymptotes are, $y = \pm \frac{4}{3} \cdot x$ $y=+-4/3*x$ .

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

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,

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

How do you write an equation for the hyperbola with asymptote y=+-(4x)/3y=±4x3y=+-(4x)/3, contains (3sqrt(2,)4)(3√2,4)(3sqrt(2,)4)?