Simplify each term in the equation in order to set the right side equal to <math><mstyle displaystyle="true"><mn>1</mn></mstyle></math> . The standard form of an ellipse or hyperbola requires the right side of the equation be <math><mstyle displaystyle="true"><mn>1</mn></mstyle></math> .

This is the form of a hyperbola. Use this form to determine the values used to find vertices and asymptotes of the hyperbola.

Match the values in this hyperbola to those of the standard form. The variable <math><mstyle displaystyle="true"><mi>h</mi></mstyle></math> represents the x-offset from the origin, <math><mstyle displaystyle="true"><mi>k</mi></mstyle></math> represents the y-offset from origin, <math><mstyle displaystyle="true"><mi>a</mi></mstyle></math> .

The center of a hyperbola follows the form of <math><mstyle displaystyle="true"><mrow><mo>(</mo><mi>h</mi><mo>,</mo><mi>k</mi><mo>)</mo></mrow></mstyle></math> . Substitute in the values of <math><mstyle displaystyle="true"><mi>h</mi></mstyle></math> and <math><mstyle displaystyle="true"><mi>k</mi></mstyle></math> .

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

Substitute the values of <math><mstyle displaystyle="true"><mi>a</mi></mstyle></math> and <math><mstyle displaystyle="true"><mi>b</mi></mstyle></math> in the formula.

Simplify.

Raise <math><mstyle displaystyle="true"><mn>3</mn></mstyle></math> to the power of <math><mstyle displaystyle="true"><mn>2</mn></mstyle></math> .

Raise <math><mstyle displaystyle="true"><mn>4</mn></mstyle></math> to the power of <math><mstyle displaystyle="true"><mn>2</mn></mstyle></math> .

Add <math><mstyle displaystyle="true"><mn>9</mn></mstyle></math> and <math><mstyle displaystyle="true"><mn>16</mn></mstyle></math> .

Rewrite <math><mstyle displaystyle="true"><mn>25</mn></mstyle></math> as <math><mstyle displaystyle="true"><msup><mrow><mn>5</mn></mrow><mrow><mn>2</mn></mrow></msup></mstyle></math> .

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

The first vertex of a hyperbola can be found by adding <math><mstyle displaystyle="true"><mi>a</mi></mstyle></math> to <math><mstyle displaystyle="true"><mi>h</mi></mstyle></math> .

Substitute the known values of <math><mstyle displaystyle="true"><mi>h</mi></mstyle></math> , <math><mstyle displaystyle="true"><mi>a</mi></mstyle></math> , and <math><mstyle displaystyle="true"><mi>k</mi></mstyle></math> into the formula and simplify.

The second vertex of a hyperbola can be found by subtracting <math><mstyle displaystyle="true"><mi>a</mi></mstyle></math> from <math><mstyle displaystyle="true"><mi>h</mi></mstyle></math> .

Substitute the known values of <math><mstyle displaystyle="true"><mi>h</mi></mstyle></math> , <math><mstyle displaystyle="true"><mi>a</mi></mstyle></math> , and <math><mstyle displaystyle="true"><mi>k</mi></mstyle></math> into the formula and simplify.

The vertices of a hyperbola follow the form of <math><mstyle displaystyle="true"><mrow><mo>(</mo><mi>h</mi><mo>±</mo><mi>a</mi><mo>,</mo><mi>k</mi><mo>)</mo></mrow></mstyle></math> . Hyperbolas have two vertices.

The first focus of a hyperbola can be found by adding <math><mstyle displaystyle="true"><mi>c</mi></mstyle></math> to <math><mstyle displaystyle="true"><mi>h</mi></mstyle></math> .

Substitute the known values of <math><mstyle displaystyle="true"><mi>h</mi></mstyle></math> , <math><mstyle displaystyle="true"><mi>c</mi></mstyle></math> , and <math><mstyle displaystyle="true"><mi>k</mi></mstyle></math> into the formula and simplify.

The second focus of a hyperbola can be found by subtracting <math><mstyle displaystyle="true"><mi>c</mi></mstyle></math> from <math><mstyle displaystyle="true"><mi>h</mi></mstyle></math> .

Substitute the known values of <math><mstyle displaystyle="true"><mi>h</mi></mstyle></math> , <math><mstyle displaystyle="true"><mi>c</mi></mstyle></math> , and <math><mstyle displaystyle="true"><mi>k</mi></mstyle></math> into the formula and simplify.

The foci of a hyperbola follow the form of <math><mstyle displaystyle="true"><mrow><mo>(</mo><mi>h</mi><mo>±</mo><msqrt><msup><mrow><mi>a</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>+</mo><msup><mrow><mi>b</mi></mrow><mrow><mn>2</mn></mrow></msup></msqrt><mo>,</mo><mi>k</mi><mo>)</mo></mrow></mstyle></math> . Hyperbolas have two foci.

Find the eccentricity by using the following formula.

Substitute the values of <math><mstyle displaystyle="true"><mi>a</mi></mstyle></math> and <math><mstyle displaystyle="true"><mi>b</mi></mstyle></math> into the formula.

Simplify the numerator.

Raise <math><mstyle displaystyle="true"><mn>3</mn></mstyle></math> to the power of <math><mstyle displaystyle="true"><mn>2</mn></mstyle></math> .

Raise <math><mstyle displaystyle="true"><mn>4</mn></mstyle></math> to the power of <math><mstyle displaystyle="true"><mn>2</mn></mstyle></math> .

Add <math><mstyle displaystyle="true"><mn>9</mn></mstyle></math> and <math><mstyle displaystyle="true"><mn>16</mn></mstyle></math> .

Rewrite <math><mstyle displaystyle="true"><mn>25</mn></mstyle></math> as <math><mstyle displaystyle="true"><msup><mrow><mn>5</mn></mrow><mrow><mn>2</mn></mrow></msup></mstyle></math> .

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

Find the value of the focal parameter of the hyperbola by using the following formula.

Substitute the values of <math><mstyle displaystyle="true"><mi>b</mi></mstyle></math> and <math><mstyle displaystyle="true"><msqrt><msup><mrow><mi>a</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>+</mo><msup><mrow><mi>b</mi></mrow><mrow><mn>2</mn></mrow></msup></msqrt></mstyle></math> in the formula.

Raise <math><mstyle displaystyle="true"><mn>4</mn></mstyle></math> to the power of <math><mstyle displaystyle="true"><mn>2</mn></mstyle></math> .

The asymptotes follow the form <math><mstyle displaystyle="true"><mi>y</mi><mo>=</mo><mo>±</mo><mfrac><mrow><mi>b</mi><mrow><mo>(</mo><mi>x</mi><mo>-</mo><mi>h</mi><mo>)</mo></mrow></mrow><mrow><mi>a</mi></mrow></mfrac><mo>+</mo><mi>k</mi></mstyle></math> because this hyperbola opens left and right.

Add <math><mstyle displaystyle="true"><mfrac><mrow><mn>4</mn></mrow><mrow><mn>3</mn></mrow></mfrac><mi>x</mi></mstyle></math> and <math><mstyle displaystyle="true"><mn>0</mn></mstyle></math> .

Combine <math><mstyle displaystyle="true"><mfrac><mrow><mn>4</mn></mrow><mrow><mn>3</mn></mrow></mfrac></mstyle></math> and <math><mstyle displaystyle="true"><mi>x</mi></mstyle></math> .

Add <math><mstyle displaystyle="true"><mo>-</mo><mfrac><mrow><mn>4</mn></mrow><mrow><mn>3</mn></mrow></mfrac><mi>x</mi></mstyle></math> and <math><mstyle displaystyle="true"><mn>0</mn></mstyle></math> .

Combine <math><mstyle displaystyle="true"><mi>x</mi></mstyle></math> and <math><mstyle displaystyle="true"><mfrac><mrow><mn>4</mn></mrow><mrow><mn>3</mn></mrow></mfrac></mstyle></math> .

Move <math><mstyle displaystyle="true"><mn>4</mn></mstyle></math> to the left of <math><mstyle displaystyle="true"><mi>x</mi></mstyle></math> .

This hyperbola has two asymptotes.

These values represent the important values for graphing and analyzing a hyperbola.

Center: <math><mstyle displaystyle="true"><mrow><mo>(</mo><mn>0</mn><mo>,</mo><mn>0</mn><mo>)</mo></mrow></mstyle></math>

Vertices: <math><mstyle displaystyle="true"><mrow><mo>(</mo><mn>3</mn><mo>,</mo><mn>0</mn><mo>)</mo></mrow><mo>,</mo><mrow><mo>(</mo><mo>-</mo><mn>3</mn><mo>,</mo><mn>0</mn><mo>)</mo></mrow></mstyle></math>

Foci: <math><mstyle displaystyle="true"><mrow><mo>(</mo><mn>5</mn><mo>,</mo><mn>0</mn><mo>)</mo></mrow><mo>,</mo><mrow><mo>(</mo><mo>-</mo><mn>5</mn><mo>,</mo><mn>0</mn><mo>)</mo></mrow></mstyle></math>

Eccentricity: <math><mstyle displaystyle="true"><mfrac><mrow><mn>5</mn></mrow><mrow><mn>3</mn></mrow></mfrac></mstyle></math>

Focal Parameter: <math><mstyle displaystyle="true"><mfrac><mrow><mn>16</mn></mrow><mrow><mn>5</mn></mrow></mfrac></mstyle></math>

Asymptotes: <math><mstyle displaystyle="true"><mi>y</mi><mo>=</mo><mfrac><mrow><mn>4</mn><mi>x</mi></mrow><mrow><mn>3</mn></mrow></mfrac></mstyle></math> , <math><mstyle displaystyle="true"><mi>y</mi><mo>=</mo><mo>-</mo><mfrac><mrow><mn>4</mn><mi>x</mi></mrow><mrow><mn>3</mn></mrow></mfrac></mstyle></math>

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Name | one hundred forty-four million eight hundred forty-eight thousand nine hundred forty-two |
---|

- 144848942 has 16 divisors, whose sum is
**262919520** - The reverse of 144848942 is
**249848441** - Previous prime number is
**17**

- Is Prime? no
- Number parity even
- Number length 9
- Sum of Digits 44
- Digital Root 8

Name | three hundred sixty-two million six hundred twenty-four thousand three hundred forty-nine |
---|

- 362624349 has 8 divisors, whose sum is
**484127712** - The reverse of 362624349 is
**943426263** - Previous prime number is
**773**

- Is Prime? no
- Number parity odd
- Number length 9
- Sum of Digits 39
- Digital Root 3

Name | four hundred eighty-two million five hundred fifty-two thousand one hundred thirty |
---|

- 482552130 has 8 divisors, whose sum is
**772083456** - The reverse of 482552130 is
**031255284** - Previous prime number is
**15**

- Is Prime? no
- Number parity even
- Number length 9
- Sum of Digits 30
- Digital Root 3