# How do you identify the vertices, foci, and direction of x^2/20-(y+1)^2/10=1?

Oct 15, 2017

When given an equation of the form:

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

The negative y term indicates a horizontal transverse axis.

The vertices are $\left(h - a , k\right)$ and $\left(h + a , k\right)$

The foci are $\left(h - \sqrt{{a}^{2} + {b}^{2}} , k\right)$ and $\left(h + \sqrt{{a}^{2} + {b}^{2}} , k\right)$

With very little work, we can write the given equation:

${x}^{2} / 20 - {\left(y + 1\right)}^{2} / 10 = 1$

In the form of equation [1]:

${\left(x - 0\right)}^{2} / {\left(2 \sqrt{5}\right)}^{2} - {\left(y - \left(- 1\right)\right)}^{2} / {\left(\sqrt{10}\right)}^{2} = 1$

The negative y term indicates a horizontal transverse axis.

The vertices are $\left(- 2 \sqrt{5} , - 1\right)$ and $\left(2 \sqrt{5} , - 1\right)$

Do the calculation for $\sqrt{{a}^{2} + {b}^{2}} = \sqrt{20 + 10} = \sqrt{30}$

The foci are $\left(- \sqrt{30} , - 1\right)$ and $\left(\sqrt{30} , - 1\right)$