# (3x^2)-(2y^2)-9x+4y-8=0 Graph and find all applicable points (center, vertex, focus, asymptote)?

## I got the equation to be ${\left(x - \frac{3}{2}\right)}^{2} / \left(\frac{12}{67}\right) - {\left(y + 1\right)}^{2} / \left(\frac{8}{67}\right)$=1 I got crazy fractions for vertex and a focus that isn't even inside the curves, so I'm pretty sure my answer is incorrect.

##### 1 Answer
Jun 15, 2016

Nothing crazy about fractions you got. But see the difference with what I got.

#### Explanation:

graph{(3x^2)-(2y^2)-9x+4y-8=0 [-10, 10, -4, 6]}
$\left(3 {x}^{2}\right) - \left(2 {y}^{2}\right) - 9 x + 4 y - 8 = 0$
Paring $x \mathmr{and} y$ terms
$\left(3 {x}^{2} - 9 x\right) - \left(2 {y}^{2} - 4 y\right) - 8 = 0$
Rearranging
$3 \left({x}^{2} - 3 x\right) - 2 \left({y}^{2} - 2 y\right) - 8 = 0$
$\implies 3 \left({x}^{2} - 2. \frac{3}{2} x\right) - 2 \left({y}^{2} - 2 y\right) - 8 = 0$
Making terms in the bracket perfect squares
$3 \left[{\left(x - \frac{3}{2}\right)}^{2} - \frac{9}{4}\right] - 2 \left[{\left(y - 1\right)}^{2} - 1\right] - 8 = 0$
Taking the constant terms out of bracket
$3 {\left(x - \frac{3}{2}\right)}^{2} - \frac{27}{4} - 2 {\left(y - 1\right)}^{2} - 2 - 8 = 0$
$\implies 3 {\left(x - \frac{3}{2}\right)}^{2} - 2 {\left(y - 1\right)}^{2} - \frac{67}{4} = 0$
Dividing both sides with $\frac{67}{4}$ and taking $1$ to RHS
$\implies \frac{3 {\left(x - \frac{3}{2}\right)}^{2}}{\frac{67}{4}} - \frac{2 {\left(y - 1\right)}^{2}}{\frac{67}{4}} = 1$
$\implies \frac{{\left(x - \frac{3}{2}\right)}^{2}}{\frac{1}{3} \times \frac{67}{4}} - {\left(y - 1\right)}^{2} / \left(\frac{1}{2} \times \frac{67}{4}\right) = 1$
$\implies \frac{{\left(x - \frac{3}{2}\right)}^{2}}{\sqrt{\frac{67}{12}}} ^ 2 - {\left(y - 1\right)}^{2} / {\left(\sqrt{\frac{67}{8}}\right)}^{2} = 1$

Comparing with general expression of hyperbola

${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$
We get
$a = \sqrt{\frac{67}{12}}$, $b = \sqrt{\frac{67}{8}}$
from values of $h \mathmr{and} k$ centre $\left(\frac{3}{2} , 1\right)$
asymptotic lines as:

$y - k = \pm \left(\frac{b}{a}\right) \left(x - h\right)$
And remaining items: foci and vertices
Cheers.