Find the Asymptotes 25x^2+36y^2-350x+325=0

$25 x^{2} + 36 y^{2} - 350 x + 325 = 0$

Step 1

Find the standard form of the ellipse.

Subtract $325$ from both sides of the equation.

$25 x^{2} + 36 y^{2} - 350 x = - 325$

Complete the square for $25 x^{2} - 350 x$ .

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

$a = 25$

$b = - 350$

$c = 0$

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 $d = \frac{b}{2 a}$ .

$d = \frac{- 350}{2 \cdot 25}$

Simplify the right side.

Cancel the common factor of $- 350$ and $2$ .

Factor $2$ out of $- 350$ .

$d = \frac{2 \cdot - 175}{2 \cdot 25}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 25$ .

$d = \frac{2 \cdot - 175}{2 (25)}$

Cancel the common factor.

$d = \frac{2 \cdot - 175}{2 \cdot 25}$

Rewrite the expression.

$d = \frac{- 175}{25}$

Cancel the common factor of $- 175$ and $25$ .

Factor $25$ out of $- 175$ .

$d = \frac{25 \cdot - 7}{25}$

Cancel the common factors.

Factor $25$ out of $25$ .

$d = \frac{25 \cdot - 7}{25 (1)}$

Cancel the common factor.

$d = \frac{25 \cdot - 7}{25 \cdot 1}$

Rewrite the expression.

$d = \frac{- 7}{1}$

Divide $- 7$ by $1$ .

$d = - 7$

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 $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{{(- 350)}^{2}}{4 \cdot 25}$

Simplify the right side.

Simplify each term.

Raise $- 350$ to the power of $2$ .

$e = 0 - \frac{122500}{4 \cdot 25}$

Multiply $4$ by $25$ .

$e = 0 - \frac{122500}{100}$

Divide $122500$ by $100$ .

$e = 0 - 1 \cdot 1225$

Multiply $- 1$ by $1225$ .

$e = 0 - 1225$

Subtract $1225$ from $0$ .

$e = - 1225$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $25 {(x - 7)}^{2} - 1225$ .

$25 {(x - 7)}^{2} - 1225$

Substitute $25 {(x - 7)}^{2} - 1225$ for $25 x^{2} - 350 x$ in the equation $25 x^{2} + 36 y^{2} - 350 x = - 325$ .

$25 {(x - 7)}^{2} - 1225 + 36 y^{2} = - 325$

Move $- 1225$ to the right side of the equation by adding $1225$ to both sides.

$25 {(x - 7)}^{2} + 36 y^{2} = - 325 + 1225$

Add $- 325$ and $1225$ .

$25 {(x - 7)}^{2} + 36 y^{2} = 900$

Divide each term by $900$ to make the right side equal to one.

$\frac{25 {(x - 7)}^{2}}{900} + \frac{36 y^{2}}{900} = \frac{900}{900}$

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 $1$ .

$\frac{{(x - 7)}^{2}}{36} + \frac{y^{2}}{25} = 1$

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.

$a = 6$

$b = 5$

$k = 0$

$h = 7$

Step 4

The center of an ellipse follows the form of $(h, k)$ . Substitute in the values of $h$ and $k$ .

$(7, 0)$

Step 5

Find $c$ , the distance from the center to a focus.

Find the distance from the center to a focus of the ellipse by using the following formula.

$\sqrt{a^{2} - b^{2}}$

Substitute the values of $a$ and $b$ in the formula.

$\sqrt{{(6)}^{2} - {(5)}^{2}}$

Simplify.

Raise $6$ to the power of $2$ .

$\sqrt{36 - {(5)}^{2}}$

Raise $5$ to the power of $2$ .

$\sqrt{36 - 1 \cdot 25}$

Multiply $- 1$ by $25$ .

$\sqrt{36 - 25}$

Subtract $25$ from $36$ .

$\sqrt{11}$

Step 6

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.

$(7 + 6, 0)$

Simplify.

$(13, 0)$

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.

$(7 - (6), 0)$

Simplify.

$(1, 0)$

Ellipses have two vertices.

${Vertex}_{1}$ : $(13, 0)$

${Vertex}_{2}$ : $(1, 0)$

${Vertex}_{1}$ : $(13, 0)$

${Vertex}_{2}$ : $(1, 0)$

Step 7

Find the foci.

The first focus of an ellipse can be found by adding $c$ to $h$ .

$(h + c, k)$

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

$(7 + \sqrt{11}, 0)$

The second focus of an ellipse can be found by subtracting $c$ from $h$ .

$(h - c, k)$

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

$(7 - (\sqrt{11}), 0)$

Simplify.

$(7 - \sqrt{11}, 0)$

Ellipses have two foci.

${Focus}_{1}$ : $(7 + \sqrt{11}, 0)$

${Focus}_{2}$ : $(7 - \sqrt{11}, 0)$

${Focus}_{1}$ : $(7 + \sqrt{11}, 0)$

${Focus}_{2}$ : $(7 - \sqrt{11}, 0)$

Step 8

Find the eccentricity.

Find the eccentricity by using the following formula.

$\frac{\sqrt{a^{2} - b^{2}}}{a}$

Substitute the values of $a$ and $b$ into the formula.

$\frac{\sqrt{{(6)}^{2} - {(5)}^{2}}}{6}$

Simplify the numerator.

Raise $6$ to the power of $2$ .

$\frac{\sqrt{36 - 5^{2}}}{6}$

Raise $5$ to the power of $2$ .

$\frac{\sqrt{36 - 1 \cdot 25}}{6}$

Multiply $- 1$ by $25$ .

$\frac{\sqrt{36 - 25}}{6}$

Subtract $25$ from $36$ .

$\frac{\sqrt{11}}{6}$

Step 9

These values represent the important values for graphing and analyzing an ellipse.

Center: $(7, 0)$

${Vertex}_{1}$ : $(13, 0)$

${Vertex}_{2}$ : $(1, 0)$

${Focus}_{1}$ : $(7 + \sqrt{11}, 0)$

${Focus}_{2}$ : $(7 - \sqrt{11}, 0)$

Eccentricity: $\frac{\sqrt{11}}{6}$

Step 10

Find the Asymptotes 25x^2+36y^2-350x+325=0

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