Find the eccentricity, centre, foci and Vertices of the hyperbola $x^{2} - 3 y^{2} + 6 x + 6 y + 18 = 0$ and draw the diagram.

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Solution

Givven, $x^{2} + 6 x - 3 y^{2} + 6 y = - 18$
$\Rightarrow (x^{2} + 6 x + 9) - 3 (y^{2} - 2 y + 1) = - 18 + 9 - 3$
$\Rightarrow {(x + 3)}^{2} - 3 {(y - 1)}^{2} = - 12$
$\Rightarrow \frac{{(x + 3)}^{2}}{- 12} - \frac{3 {(y - 1)}^{2}}{- 12} = 1$
$\Rightarrow \frac{{(y - 1)}^{2}}{4} - \frac{{(x + 3)}^{2}}{12} = 1$
$\Rightarrow \frac{y^{2}}{4} - \frac{x^{2}}{12} = 1$ where $\Rightarrow X = x + 3$ ; $Y = y - 1$
$\Rightarrow a^{2} = 4, b^{2} = 12$
$\Rightarrow e = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1 + \frac{12}{4}} = 2$
Referred to $X$ , $Y$
Referred to $x$ , $y$
$X = x + 3$ , $Y = y - 1$
Centre
$(0, 0)$
$X = 0$ , $Y = 0$
$x + 3 = 0 \Rightarrow x = - 3$
$y - 1 = 0 \Rightarrow y = 1$
$X (- 3, 1)$
Foci
$(0, a e) (0, - a e)$
$(0, 4) (0, - 4)$ $(- 3, - 3) (- 3, 5)$
Vertices
$(0, a) (0, - a)$
$(0, 2) (0, - 2)$ $(- 3, 3) (- 3, - 1)$

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	Referred to $X$ , $Y$	Referred to $x$ , $y$ $X = x + 3$ , $Y = y - 1$
Centre	$(0, 0)$	$X = 0$ , $Y = 0$ $x + 3 = 0 \Rightarrow x = - 3$ $y - 1 = 0 \Rightarrow y = 1$ $X (- 3, 1)$
Foci	$(0, a e) (0, - a e)$ $(0, 4) (0, - 4)$	$(- 3, - 3) (- 3, 5)$
Vertices	$(0, a) (0, - a)$ $(0, 2) (0, - 2)$	$(- 3, 3) (- 3, - 1)$

Find the eccentricity, centre, foci and Vertices of the hyperbola x2−3y2+6x+6y+18=0 and draw the diagram.

Find the eccentricity, centre, foci and Vertices of the hyperbola $x^{2} - 3 y^{2} + 6 x + 6 y + 18 = 0$ and draw the diagram.