Find the Vertices 25y^2-16x^2=400

$25 y^{2} - 16 x^{2} = 400$

Step 1

Find the standard form of the hyperbola.

Divide each term by $400$ to make the right side equal to one.

$\frac{25 y^{2}}{400} - \frac{16 x^{2}}{400} = \frac{400}{400}$

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 $1$ .

$\frac{y^{2}}{16} - \frac{x^{2}}{25} = 1$

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, $a$ .

$a = 4$

$b = 5$

$k = 0$

$h = 0$

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.

$(0, 4)$

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.

$(0, - 4)$

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

$(0, 4), (0, - 4)$

Step 5

Find the Vertices 25y^2-16x^2=400

Related Questions

Support LCMGCF.com by disabling your adblocker.