Find the Vertices 16y^2-4x^2=64

$16 y^{2} - 4 x^{2} = 64$

Step 1

Find the standard form of the hyperbola.

Divide each term by $64$ to make the right side equal to one.

$\frac{16 y^{2}}{64} - \frac{4 x^{2}}{64} = \frac{64}{64}$

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 $1$ .

$\frac{y^{2}}{4} - \frac{x^{2}}{16} = 1$

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.

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

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, $a$ .

$a = 2$

$b = 4$

$k = 0$

$h = 0$

Step 4

Find the vertices.

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

$(h, k + a)$

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

$(0, 2)$

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

$(h, k - a)$

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

$(0, - 2)$

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

$(0, 2), (0, - 2)$

Step 5

Find the Vertices 16y^2-4x^2=64

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