Algebra Examples

$(0, 0)$ , $(4, 0)$ , $(6, 0)$

Step 1

There are two general equations for an ellipse.

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

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

Step 2

$a$ is the distance between the vertex $(6, 0)$ and the center point $(0, 0)$ .

Tap for more steps...

Step 2.1

Use the distance formula to determine the distance between the two points.

$Distance = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

Step 2.2

Substitute the actual values of the points into the distance formula.

$a = \sqrt{{(6 - 0)}^{2} + {(0 - 0)}^{2}}$

Step 2.3

Simplify.

Tap for more steps...

Step 2.3.1

Subtract $0$ from $6$ .

$a = \sqrt{6^{2} + {(0 - 0)}^{2}}$

Step 2.3.2

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

$a = \sqrt{36 + {(0 - 0)}^{2}}$

Step 2.3.3

Subtract $0$ from $0$ .

$a = \sqrt{36 + 0^{2}}$

Step 2.3.4

Raising $0$ to any positive power yields $0$ .

$a = \sqrt{36 + 0}$

Step 2.3.5

Add $36$ and $0$ .

$a = \sqrt{36}$

Step 2.3.6

Rewrite $36$ as $6^{2}$ .

$a = \sqrt{6^{2}}$

Step 2.3.7

Pull terms out from under the radical, assuming positive real numbers.

$a = 6$

Step 3

$c$ is the distance between the focus $(4, 0)$ and the center $(0, 0)$ .

Tap for more steps...

Step 3.1

Use the distance formula to determine the distance between the two points.

$Distance = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

Step 3.2

Substitute the actual values of the points into the distance formula.

$c = \sqrt{{(4 - 0)}^{2} + {(0 - 0)}^{2}}$

Step 3.3

Simplify.

Tap for more steps...

Step 3.3.1

Subtract $0$ from $4$ .

$c = \sqrt{4^{2} + {(0 - 0)}^{2}}$

Step 3.3.2

Raise $4$ to the power of $2$ .

$c = \sqrt{16 + {(0 - 0)}^{2}}$

Step 3.3.3

Subtract $0$ from $0$ .

$c = \sqrt{16 + 0^{2}}$

Step 3.3.4

Raising $0$ to any positive power yields $0$ .

$c = \sqrt{16 + 0}$

Step 3.3.5

Add $16$ and $0$ .

$c = \sqrt{16}$

Step 3.3.6

Rewrite $16$ as $4^{2}$ .

$c = \sqrt{4^{2}}$

Step 3.3.7

Pull terms out from under the radical, assuming positive real numbers.

$c = 4$

Step 4

Using the equation $c^{2} = a^{2} - b^{2}$ . Substitute $6$ for $a$ and $4$ for $c$ .

Tap for more steps...

Step 4.1

Rewrite the equation as ${(6)}^{2} - b^{2} = 4^{2}$ .

${(6)}^{2} - b^{2} = 4^{2}$

Step 4.2

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

$36 - b^{2} = 4^{2}$

Step 4.3

Raise $4$ to the power of $2$ .

$36 - b^{2} = 16$

Step 4.4

Move all terms not containing $b$ to the right side of the equation.

Tap for more steps...

Step 4.4.1

Subtract $36$ from both sides of the equation.

$- b^{2} = 16 - 36$

Step 4.4.2

Subtract $36$ from $16$ .

$- b^{2} = - 20$

Step 4.5

Divide each term in $- b^{2} = - 20$ by $- 1$ and simplify.

Tap for more steps...

Step 4.5.1

Divide each term in $- b^{2} = - 20$ by $- 1$ .

$\frac{- b^{2}}{- 1} = \frac{- 20}{- 1}$

Step 4.5.2

Simplify the left side.

Tap for more steps...

Step 4.5.2.1

Dividing two negative values results in a positive value.

$\frac{b^{2}}{1} = \frac{- 20}{- 1}$

Step 4.5.2.2

Divide $b^{2}$ by $1$ .

$b^{2} = \frac{- 20}{- 1}$

Step 4.5.3

Simplify the right side.

Tap for more steps...

Step 4.5.3.1

Divide $- 20$ by $- 1$ .

$b^{2} = 20$

Step 4.6

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$b = \pm \sqrt{20}$

Step 4.7

Simplify $\pm \sqrt{20}$ .

Tap for more steps...

Step 4.7.1

Rewrite $20$ as $2^{2} \cdot 5$ .

Tap for more steps...

Step 4.7.1.1

Factor $4$ out of $20$ .

$b = \pm \sqrt{4 (5)}$

Step 4.7.1.2

Rewrite $4$ as $2^{2}$ .

$b = \pm \sqrt{2^{2} \cdot 5}$

Step 4.7.2

Pull terms out from under the radical.

$b = \pm 2 \sqrt{5}$

Step 4.8

The complete solution is the result of both the positive and negative portions of the solution.

Tap for more steps...

Step 4.8.1

First, use the positive value of the $\pm$ to find the first solution.

$b = 2 \sqrt{5}$

Step 4.8.2

Next, use the negative value of the $\pm$ to find the second solution.

$b = - 2 \sqrt{5}$

Step 4.8.3

The complete solution is the result of both the positive and negative portions of the solution.

$b = 2 \sqrt{5}, - 2 \sqrt{5}$

Step 5

$b$ is a distance, which means it should be a positive number.

$b = 2 \sqrt{5}$

Step 6

The slope of the line between the focus $(4, 0)$ and the center $(0, 0)$ determines whether the ellipse is vertical or horizontal. If the slope is $0$ , the graph is horizontal. If the slope is undefined, the graph is vertical.

Tap for more steps...

Step 6.1

Slope is equal to the change in $y$ over the change in $x$ , or rise over run.

$m = \frac{change in y}{change in x}$

Step 6.2

The change in $x$ is equal to the difference in x-coordinates (also called run), and the change in $y$ is equal to the difference in y-coordinates (also called rise).

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

Step 6.3

Substitute in the values of $x$ and $y$ into the equation to find the slope.

$m = \frac{0 - (0)}{0 - (4)}$

Step 6.4

Simplify.

Tap for more steps...

Step 6.4.1

Simplify the numerator.

Tap for more steps...

Step 6.4.1.1

Multiply $- 1$ by $0$ .

$m = \frac{0 + 0}{0 - (4)}$

Step 6.4.1.2

Add $0$ and $0$ .

$m = \frac{0}{0 - (4)}$

Step 6.4.2

Simplify the denominator.

Tap for more steps...

Step 6.4.2.1

Multiply $- 1$ by $4$ .

$m = \frac{0}{0 - 4}$

Step 6.4.2.2

Subtract $4$ from $0$ .

$m = \frac{0}{- 4}$

Step 6.4.3

Divide $0$ by $- 4$ .

$m = 0$

Step 6.5

The general equation for a horizontal ellipse is $\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1$ .

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

Step 7

Substitute the values $h = 0$ , $k = 0$ , $a = 6$ , and $b = 2 \sqrt{5}$ into $\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1$ to get the ellipse equation $\frac{{(x - (0))}^{2}}{{(6)}^{2}} + \frac{{(y - (0))}^{2}}{{(2 \sqrt{5})}^{2}} = 1$ .

$\frac{{(x - (0))}^{2}}{{(6)}^{2}} + \frac{{(y - (0))}^{2}}{{(2 \sqrt{5})}^{2}} = 1$

Step 8

Simplify to find the final equation of the ellipse.

Tap for more steps...

Step 8.1

Simplify the numerator.

Tap for more steps...

Step 8.1.1

Multiply $- 1$ by $0$ .

$\frac{{(x + 0)}^{2}}{6^{2}} + \frac{{(y - (0))}^{2}}{{(2 \sqrt{5})}^{2}} = 1$

Step 8.1.2

Add $x$ and $0$ .

$\frac{x^{2}}{6^{2}} + \frac{{(y - (0))}^{2}}{{(2 \sqrt{5})}^{2}} = 1$

Step 8.2

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

$\frac{x^{2}}{36} + \frac{{(y - (0))}^{2}}{{(2 \sqrt{5})}^{2}} = 1$

Step 8.3

Simplify the numerator.

Tap for more steps...

Step 8.3.1

Multiply $- 1$ by $0$ .

$\frac{x^{2}}{36} + \frac{{(y + 0)}^{2}}{{(2 \sqrt{5})}^{2}} = 1$

Step 8.3.2

Add $y$ and $0$ .

$\frac{x^{2}}{36} + \frac{y^{2}}{{(2 \sqrt{5})}^{2}} = 1$

Step 8.4

Simplify the denominator.

Tap for more steps...

Step 8.4.1

Apply the product rule to $2 \sqrt{5}$ .

$\frac{x^{2}}{36} + \frac{y^{2}}{2^{2} {\sqrt{5}}^{2}} = 1$

Step 8.4.2

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

$\frac{x^{2}}{36} + \frac{y^{2}}{4 {\sqrt{5}}^{2}} = 1$

Step 8.4.3

Rewrite ${\sqrt{5}}^{2}$ as $5$ .

Tap for more steps...

Step 8.4.3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5}$ as $5^{\frac{1}{2}}$ .

$\frac{x^{2}}{36} + \frac{y^{2}}{4 {(5^{\frac{1}{2}})}^{2}} = 1$

Step 8.4.3.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{x^{2}}{36} + \frac{y^{2}}{4 \cdot 5^{\frac{1}{2} \cdot 2}} = 1$

Step 8.4.3.3

Combine $\frac{1}{2}$ and $2$ .

$\frac{x^{2}}{36} + \frac{y^{2}}{4 \cdot 5^{\frac{2}{2}}} = 1$

Step 8.4.3.4

Cancel the common factor of $2$ .

Tap for more steps...

Step 8.4.3.4.1

Cancel the common factor.

$\frac{x^{2}}{36} + \frac{y^{2}}{4 \cdot 5^{\frac{2}{2}}} = 1$

Step 8.4.3.4.2

Rewrite the expression.

$\frac{x^{2}}{36} + \frac{y^{2}}{4 \cdot 5} = 1$

Step 8.4.3.5

Evaluate the exponent.

$\frac{x^{2}}{36} + \frac{y^{2}}{4 \cdot 5} = 1$

Step 8.5

Multiply $4$ by $5$ .

$\frac{x^{2}}{36} + \frac{y^{2}}{20} = 1$

Step 9

Enter YOUR Problem