Find the equations of directrices, latus rectums and length of latus rectum for the hyperbola $9x^2 - 4y^2 - 36x + 32y + 8 = 0$
Equation of given hyperbola is: $\;$ $9x^2 - 4y^2 - 36x + 32y + 8 = 0$
i.e. $\;$ $9 \left(x^2 - 4x\right) - 4 \left(y^2 - 8y\right) = -8$
i.e. $\;$ $9 \left[\left(x^2 - 4x + 4\right) - 4\right] - 4 \left[\left(y^2 - 8y + 16\right) - 16\right] = -8$
i.e. $\;$ $9 \left(x - 2\right)^2 - 4 \left(y - 4\right)^2 = 36 - 64 - 8$
i.e. $\;$ $9 \left(x - 2\right)^2 - 4 \left(y - 4\right)^2 = -36$
i.e. $\;$ $\dfrac{\left(y - 4\right)^2}{36 /4} - \dfrac{\left(x - 2\right)^2}{36 / 9} = 1$
i.e. $\;$ $\dfrac{\left(y - 4\right)^2}{9} - \dfrac{\left(x - 2\right)^2}{4} = 1$ $\;\;\; \cdots \; (1)$
Let $Y = y - 4$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $X = x - 2$ $\;\;\; \cdots \; (2b)$
$\therefore$ $\;$ In view of equations $(2a)$ and $(2b)$ equation $(1)$ becomes
$\dfrac{Y^2}{9} - \dfrac{X^2}{4} = 1$ $\;\;\; \cdots \; (3)$
For the hyperbola given by equation $(3)$, the transverse axis 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
$a^2 = 9 \implies a = 3$ $\;$ and $\;$ $b^2 = 4 \implies b = 2$
Now, $e = \sqrt{1 + \dfrac{b^2}{a^2}} = \sqrt{1 + \dfrac{4}{9}} = \dfrac{\sqrt{13}}{3}$
| Referred to X, Y | Referred to x, y $X = x - 2$; $\;$ $Y = y - 4$ i.e. $\;$ $x = X + 2$; $\;$ $y = Y + 4$ |
|
|---|---|---|
| Equations of directrices | $Y = \pm \dfrac{a}{e} = \pm \dfrac{3}{\sqrt{13} / 3}$ i.e. $\;$ $Y = \pm \dfrac{9}{\sqrt{13}}$ |
$Y = \pm \dfrac{9}{\sqrt{13}}$ $ \implies y = 4 \pm \dfrac{9}{\sqrt{13}}$ |
| Equations of latus rectum | $Y = \pm a \cdot e = \pm 3 \times \dfrac{\sqrt{13}}{3}$ i.e. $\;$ $Y = \pm \sqrt{13}$ |
$Y = \pm \sqrt{13}$ $\implies$ $y = 4 \pm \sqrt{13}$ |
| Length of latus rectum | $\dfrac{2b^2}{a} = \dfrac{2 \times 4}{3} = \dfrac{8}{3}$ | $\dfrac{8}{3}$ |