Find the Vertices x^2+4x+16y^2-96y+4=0

$x^{2} + 4 x + 16 y^{2} - 96 y + 4 = 0$

Step 1

Find the standard form of the ellipse.

Subtract $4$ from both sides of the equation.

$x^{2} + 4 x + 16 y^{2} - 96 y = - 4$

Complete the square for $x^{2} + 4 x$ .

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 1$

$b = 4$

$c = 0$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{4}{2 \cdot 1}$

Cancel the common factor of $4$ and $2$ .

Factor $2$ out of $4$ .

$d = \frac{2 \cdot 2}{2 \cdot 1}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 1$ .

$d = \frac{2 \cdot 2}{2 (1)}$

Cancel the common factor.

$d = \frac{2 \cdot 2}{2 \cdot 1}$

Rewrite the expression.

$d = \frac{2}{1}$

Divide $2$ by $1$ .

$d = 2$

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{4^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $4^{2}$ and $4$ .

Factor $4$ out of $4^{2}$ .

$e = 0 - \frac{4 \cdot 4}{4 \cdot 1}$

Cancel the common factors.

Factor $4$ out of $4 \cdot 1$ .

$e = 0 - \frac{4 \cdot 4}{4 (1)}$

Cancel the common factor.

$e = 0 - \frac{4 \cdot 4}{4 \cdot 1}$

Rewrite the expression.

$e = 0 - \frac{4}{1}$

Divide $4$ by $1$ .

$e = 0 - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$e = 0 - 4$

Subtract $4$ from $0$ .

$e = - 4$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x + 2)}^{2} - 4$ .

${(x + 2)}^{2} - 4$

Substitute ${(x + 2)}^{2} - 4$ for $x^{2} + 4 x$ in the equation $x^{2} + 4 x + 16 y^{2} - 96 y = - 4$ .

${(x + 2)}^{2} - 4 + 16 y^{2} - 96 y = - 4$

Move $- 4$ to the right side of the equation by adding $4$ to both sides.

${(x + 2)}^{2} + 16 y^{2} - 96 y = - 4 + 4$

Complete the square for $16 y^{2} - 96 y$ .

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 16$

$b = - 96$

$c = 0$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{- 96}{2 \cdot 16}$

Simplify the right side.

Cancel the common factor of $- 96$ and $2$ .

Factor $2$ out of $- 96$ .

$d = \frac{2 \cdot - 48}{2 \cdot 16}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 16$ .

$d = \frac{2 \cdot - 48}{2 (16)}$

Cancel the common factor.

$d = \frac{2 \cdot - 48}{2 \cdot 16}$

Rewrite the expression.

$d = \frac{- 48}{16}$

Cancel the common factor of $- 48$ and $16$ .

Factor $16$ out of $- 48$ .

$d = \frac{16 \cdot - 3}{16}$

Cancel the common factors.

Factor $16$ out of $16$ .

$d = \frac{16 \cdot - 3}{16 (1)}$

Cancel the common factor.

$d = \frac{16 \cdot - 3}{16 \cdot 1}$

Rewrite the expression.

$d = \frac{- 3}{1}$

Divide $- 3$ by $1$ .

$d = - 3$

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{{(- 96)}^{2}}{4 \cdot 16}$

Simplify the right side.

Simplify each term.

Raise $- 96$ to the power of $2$ .

$e = 0 - \frac{9216}{4 \cdot 16}$

Multiply $4$ by $16$ .

$e = 0 - \frac{9216}{64}$

Divide $9216$ by $64$ .

$e = 0 - 1 \cdot 144$

Multiply $- 1$ by $144$ .

$e = 0 - 144$

Subtract $144$ from $0$ .

$e = - 144$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $16 {(y - 3)}^{2} - 144$ .

$16 {(y - 3)}^{2} - 144$

Substitute $16 {(y - 3)}^{2} - 144$ for $16 y^{2} - 96 y$ in the equation $x^{2} + 4 x + 16 y^{2} - 96 y = - 4$ .

${(x + 2)}^{2} + 16 {(y - 3)}^{2} - 144 = - 4 + 4$

Move $- 144$ to the right side of the equation by adding $144$ to both sides.

${(x + 2)}^{2} + 16 {(y - 3)}^{2} = - 4 + 4 + 144$

Simplify $- 4 + 4 + 144$ .

Add $- 4$ and $4$ .

${(x + 2)}^{2} + 16 {(y - 3)}^{2} = 0 + 144$

Add $0$ and $144$ .

${(x + 2)}^{2} + 16 {(y - 3)}^{2} = 144$

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

$\frac{{(x + 2)}^{2}}{144} + \frac{16 {(y - 3)}^{2}}{144} = \frac{144}{144}$

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 $1$ .

$\frac{{(x + 2)}^{2}}{144} + \frac{{(y - 3)}^{2}}{9} = 1$

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.

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

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.

$a = 12$

$b = 3$

$k = 3$

$h = - 2$

Step 4

Find the vertices.

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

$(h + a, k)$

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

$(- 2 + 12, 3)$

Simplify.

$(10, 3)$

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

$(h - a, k)$

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

$(- 2 - (12), 3)$

Simplify.

$(- 14, 3)$

Ellipses have two vertices.

${Vertex}_{1}$ : $(10, 3)$

${Vertex}_{2}$ : $(- 14, 3)$

${Vertex}_{1}$ : $(10, 3)$

${Vertex}_{2}$ : $(- 14, 3)$

Step 5

Find the Vertices x^2+4x+16y^2-96y+4=0

Related Questions

Support LCMGCF.com by disabling your adblocker.