Find the Vertices 16(x-2)^2-9(y-1)^2=144

$16 {(x - 2)}^{2} - 9 {(y - 1)}^{2} = 144$

Step 1

Find the standard form of the hyperbola.

Divide each term by $144$ to make the right side equal to one.

$\frac{16 {(x - 2)}^{2}}{144} - \frac{9 {(y - 1)}^{2}}{144} = \frac{144}{144}$

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 $1$ .

$\frac{{(x - 2)}^{2}}{9} - \frac{{(y - 1)}^{2}}{16} = 1$

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, $a$ .

$a = 3$

$b = 4$

$k = 1$

$h = 2$

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.

$(5, 1)$

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.

$(- 1, 1)$

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

$(5, 1), (- 1, 1)$

Step 5

Find the Vertices 16(x-2)^2-9(y-1)^2=144

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