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

Jan 25, 2018

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

#### Explanation:

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, $y = \pm \frac{4}{3} \cdot x$.

$\therefore \frac{b}{a} = \frac{4}{3.} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(1\right)$.

Next, $\left(3 \sqrt{2} , 4\right) \in S \Rightarrow {\left(3 \sqrt{2}\right)}^{2} / {a}^{2} - {4}^{2} / {b}^{2} = 1$.

$\therefore \frac{18}{a} ^ 2 - \frac{16}{b} ^ 2 = 1$.

Multiplying by ${b}^{2}$, we get, $18 \cdot {b}^{2} / {a}^{2} - 16 = {b}^{2}$.

By $\left(1\right)$, then, $18 \cdot {\left(\frac{4}{3}\right)}^{2} - 16 = {b}^{2} , i . e . , {b}^{2} = 16$.

$\therefore b = 4 \left(\because , b > 0\right) , \mathmr{and} , \left(1\right) \Rightarrow a = 3 , \mathmr{and} , {a}^{2} = 9$.

With ${a}^{2} = 9 , \mathmr{and} {b}^{2} = 16$, the reqd. eqn. of the Hyperbola is,

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