Find the eccentricity, center, foci and vertices of the hyperbola $x^2 - 3y^2 + 6x + 6y + 18 = 0$ and sketch it.
Equation of given hyperbola is
$x^2 - 3y^2 + 6x + 6y + 18 = 0$
i.e. $\left(x^2 + 6x\right) - 3 \left(y^2 - 2y\right) = -18$
i.e. $\left(x^2 + 6x + 9\right) - 9 - 3 \left[\left(y^2 - 2y + 1\right) - 1\right] = -18$
i.e. $\left(x + 3\right)^2 - 3 \left(y - 1\right)^2 = -18 + 9 - 3$
i.e. $\left(x + 3\right)^2 - 3 \left(y - 1\right)^2 = -12$
i.e. $\dfrac{\left(y - 1\right)^2}{12 / 3} - \dfrac{\left(x + 3\right)^2}{12} = 1$
i.e. $\dfrac{\left(y - 1\right)^2}{4} - \dfrac{\left(x + 3\right)^2}{12} = 1$ $\;\;\; \cdots \; (1)$
Let $\;$ $Y = y - 1$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $X = x + 3$ $\;\;\; \cdots \; (2b)$
Then, we have from equations $(1)$, $(2a)$ and $(2b)$,
$\dfrac{Y^2}{4} - \dfrac{X^2}{12} = 1$ $\;\;\; \cdots \; (3)$
The transverse axis of the hyperbola given by equation $(3)$ is along the Y axis.
Comparing equation $(3)$ with the standard equation of hyperbola $\;$ $\dfrac{Y^2}{a^2} - \dfrac{X^2}{b^2} = 1$ $\;$ gives
$a^2 = 4 \implies a = 2$ $\;$ and $\;$ $b^2 = 12 \implies b = 2 \sqrt{3}$
Eccentricity $= e = \sqrt{1 + \dfrac{b^2}{a^2}} = \sqrt{1 + \dfrac{12}{4}} = \sqrt{4} = 2$
| Referred to X, Y | Referred to x, y $X = x + 3$; $\;$ $Y = y - 1$ i.e. $\;$ $x = X - 3$; $\;$ $y = Y + 1$ |
|
|---|---|---|
| Center | $C \left(0, 0\right)$ | $X = 0 \implies x = -3$ $Y = 0 \implies y = 1$ $\therefore$ $\;$ Center $C' = \left(-3, 1\right)$ |
| Foci | $F_1 = \left(0, ae\right) = \left(0, 2 \times 2\right)$ i.e. $\;$ $F_1 = \left(0,4\right)$ |
$X = 0 \implies x = -3$ $Y = 4 \implies y = 5$ $\therefore$ $\;$ $F'_1 = \left(-3, 5\right)$ |
| $F_2 = \left(0, -ae\right) = \left(0, -2 \times 2\right)$ i.e. $\;$ $F_2 = \left(0, -4\right)$ |
$X = 0 \implies x = -3$ $Y = -4 \implies y = -3$ $\therefore$ $\;$ $F'_2 = \left(-3, -3\right)$ |
|
| Vertices | $A_1 = \left(0, a\right) = \left(0, 2\right)$ | $X = 0 \implies x = -3$ $Y = 2 \implies y = 3$ $\therefore$ $\;$ $A'_1 = \left(-3, 3\right)$ |
| $A_2 = \left(0, -a\right) = \left(0, -2\right)$ | $X = 0 \implies x = -3$ $Y = -2 \implies y = -1$ $\therefore$ $\;$ $A'_2 = \left(-3, -1\right)$ |
