Answer to Question #162061 in Analytic Geometry for JMD

Question

 Draw the hyperbola with the equation: 3𝑥 2 − 2𝑦 2 + 6𝑥 − 8𝑦 = 6

Answer 1

first of all, we need to transform the equation to the canonical form

$\boxed{\frac {(x-x_0)^2} {a^2} - \frac{(y-y_0)^2} {b^2} = 1}$ , $x_0$ and $y_0$ are coordinates of the center,

$3x^{2}+6x-2y^{2}-8y=6$

$(3x^2 + 6x + 3) - (2y^2 +8y+ 8) = 1$

$\dfrac{(x+1)^2}{\dfrac1 3} - \dfrac{(y+2)^2}{\dfrac 1 2} = 1$

O (-1,-2) is the center

$a = \sqrt{\dfrac1 3} \thickapprox 0.6$

$b = \sqrt{\dfrac1 2} \thickapprox 0.7$

what is a and b:

(instead of the origin, there can be any point that is the center of the hyperbola)