Find the Vertices 4x^2+y^2-36=0

$4 x^{2} + y^{2} - 36 = 0$

Step 1

Find the standard form of the ellipse.

Add $36$ to both sides of the equation.

$4 x^{2} + y^{2} = 36$

Divide each term by $36$ to make the right side equal to one.

$\frac{4 x^{2}}{36} + \frac{y^{2}}{36} = \frac{36}{36}$

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 $1$ .

$\frac{x^{2}}{9} + \frac{y^{2}}{36} = 1$

Step 2

This is the form of an ellipse. Use this form to determine the values used to find the center along with the major and minor axis of the ellipse.

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

Step 3

Match the values in this ellipse to those of the standard form. The variable $a$ represents the radius of the major axis of the ellipse, $b$ represents the radius of the minor axis of the ellipse, $h$ represents the x-offset from the origin, and $k$ represents the y-offset from the origin.

$a = 6$

$b = 3$

$k = 0$

$h = 0$

Step 4

Find the vertices.

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

$(h, k + a)$

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

$(0, 0 + 6)$

Simplify.

$(0, 6)$

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

$(h, k - a)$

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

$(0, 0 - (6))$

Simplify.

$(0, - 6)$

Ellipses have two vertices.

${Vertex}_{1}$ : $(0, 6)$

${Vertex}_{2}$ : $(0, - 6)$

${Vertex}_{1}$ : $(0, 6)$

${Vertex}_{2}$ : $(0, - 6)$

Step 5

Find the Vertices 4x^2+y^2-36=0

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