How do you identify the vertices, foci, and direction of (x-1)^2/4-(y-3)^2/4=1?

Nov 8, 2016

The vertices are $\left(3 , 3\right)$ and $\left(- 1 , 3\right)$
The foci are $F \left(1 + 2 \sqrt{2} , 3\right)$ and $F ' \left(1 - 2 \sqrt{2} , 3\right)$
The asymptotes are $y = x + 2$ and $y = - x + 4$

Explanation:

This ia a left right hyperbola.
${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$
The center is $\left(h , k\right) = \left(1 , 3\right)$
The vertices are $\left(h + a , k\right) = \left(3 , 3\right)$ and $\left(h - a , k\right) = \left(- 1 , 3\right)$
The slope of the asymptotes are $\pm \frac{b}{a} = \pm 1$
The equations of the asymptotes are $y = k \pm \frac{b}{a} \left(x - h\right)$
$\implies$ $y = 3 + 1 \left(x - 1\right) = x + 2$
and $y = 3 - \left(x - 1\right) = - x + 4$
To calculate the foci, we need $c = \pm \sqrt{{a}^{2} + {b}^{2}}$
$\implies$ $c = \pm \sqrt{8} = \pm 2 \sqrt{2}$
the foci are $h \pm c , k$
$\implies$ $F \left(1 + 2 \sqrt{2} , 3\right)$ and $F ' \left(1 - 2 \sqrt{2} , 3\right)$
graph{((x-1)^2/4-(y-3)^2/4-1)(y-x-2)(y+x-4)=0 [-10, 10, -2, 8]}