# How do you find the vertex, focus, and directrix of a hyperbola ((x+3)^2/4) - ((y-4)^2/25)=1?

Nov 21, 2016

Vertices are $\left(- 1 , 4\right)$ and $\left(- 5 , 4\right)$; focii are $\left(- 3 + \sqrt{29} , 4\right)$ and $\left(- 3 - \sqrt{29} , 4\right)$ and directrix are $x = - \frac{4}{\sqrt{29}}$ and $x = \frac{4}{\sqrt{29}}$

#### Explanation:

Here the equation of hyperbola is

${\left(x + 3\right)}^{2} / 4 - {\left(y - 4\right)}^{2} / 25 = 1$ or ${\left(x + 3\right)}^{2} / {2}^{2} - {\left(y - 4\right)}^{2} / {5}^{2} = 1$

If the general equation of hyperbola is

${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$, then

vertices are $\left(h \pm a , k\right)$, focii are $\left(h \pm c , k\right)$, where ${c}^{2} = {a}^{2} + {b}^{2}$

and directrix are $x = \pm {a}^{2} / c$

Here $a = 2$, $b = 5$, $c = \sqrt{{2}^{2} + {5}^{2}} = \sqrt{29}$ and $h = - 3$, $k = 4$

Hence for hyperbola ${\left(x + 3\right)}^{2} / {2}^{2} - {\left(y - 4\right)}^{2} / {5}^{2} = 1$

Vertices are $\left(- 3 \pm 2 , 4\right)$ i.e. $\left(- 1 , 4\right)$ and $\left(- 5 , 4\right)$

Focii are $\left(- 3 + \sqrt{29} , 4\right)$ and $\left(- 3 - \sqrt{29} , 4\right)$

and directrix are $x = - \frac{4}{\sqrt{29}}$ and $x = \frac{4}{\sqrt{29}}$
graph{(x+3)^2/2^2-(y-4)^2/5^2=1 [-24.42, 15.58, -5.68, 14.32]}