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

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Solution

Given equation of hyperbola is $x^{2} - 3 y^{2} + 6 x + 6 y + 18 = 0$
$(x^{2} + 6 x) - (3 y^{2} - 6 y) = - 18$
$(x^{2} + 6 x + 9 - 9) - 3 (y^{2} - 2 y + 1 - 1) = - 18$
${(x + 3)}^{2} - 9 - 3 {(y - 1)}^{2} + 3 = - 18$
$3 {(y - 1)}^{2} - {(x + 3)}^{2} = 12$
( $\div$ by $12$ ) $\frac{{(y - 1)}^{2}}{4} - \frac{{(x + 3)}^{2}}{12} = 1$ ; Centre $(- 3, 1)$
Transverse axis is parallel to $y$ -axis
$a^{2} = 4, b^{2} = 12$
$e = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1 + \frac{12}{4}} = \sqrt{1 + 3} = 2$
$a = 2, e = 2$
$a e = 2 \times 2 = 4; e = 2$
Centre $= (- 3, 1)$ ; Foci $= (0, \pm a e) = (0, \pm 4)$
(i.e.,) $F_{1} = (0, 4) + (- 3, 1) = (- 3, 5)$
$F_{2} = (0, - 4) + (- 3, 1) = (- 3, - 5)$
Vertices $(0, \pm a) = (0, \pm 2)$
(i.e.,) $V_{1} = (0, 2) + (- 3, 1) = (- 3, 3)$
and $V_{2} = (0, - 2) + (- 3, 1) = (- 3, - 1)$

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

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