Find the Vertices ((x+8)^2)/16-(y+4)^2=1

$\frac{{(x + 8)}^{2}}{16} - {(y + 4)}^{2} = 1$

Step 1

Simplify each term in the equation in order to set the right side equal to $1$ . The standard form of an ellipse or hyperbola requires the right side of the equation be $1$ .

$\frac{{(x + 8)}^{2}}{16} - {(y + 4)}^{2} = 1$

Step 2

This is the form of a hyperbola. Use this form to determine the values used to find vertices and asymptotes of the hyperbola.

$\frac{{(x - h)}^{2}}{a^{2}} - \frac{{(y - k)}^{2}}{b^{2}} = 1$

Step 3

Match the values in this hyperbola to those of the standard form. The variable $h$ represents the x-offset from the origin, $k$ represents the y-offset from origin, $a$ .

$a = 4$

$b = 1$

$k = - 4$

$h = - 8$

Step 4

Find the vertices.

The first vertex of a hyperbola can be found by adding $a$ to $h$ .

$(h + a, k)$

Substitute the known values of $h$ , $a$ , and $k$ into the formula and simplify.

$(- 4, - 4)$

The second vertex of a hyperbola can be found by subtracting $a$ from $h$ .

$(h - a, k)$

Substitute the known values of $h$ , $a$ , and $k$ into the formula and simplify.

$(- 12, - 4)$

The vertices of a hyperbola follow the form of $(h \pm a, k)$ . Hyperbolas have two vertices.

$(- 4, - 4), (- 12, - 4)$

Step 5

Find the Vertices ((x+8)^2)/16-(y+4)^2=1

Related Questions

Support LCMGCF.com by disabling your adblocker.