QuestionGraph the hyperbola $\frac{y^{2}}{16}-x^{2}=1$ and determine its asymptotes.

Studdy Solution

STEP 1

Assumptions1. The given equation is of a hyperbola. The equation is in the standard form $\frac{y^{}}{a^{}} - \frac{x^{}}{b^{}} =1$
3. The center of the hyperbola is at the origin (0,0)

STEP 2

From the given equation, we can identify the values of $a^{2}$ and $b^{2}$ . Here, $a^{2} =16$ and $b^{2} =1$ .

STEP 3

To find the values of a and b, we take the square root of $a^{2}$ and $b^{2}$ .
$a = \sqrt{16} =$ $b = \sqrt{1} =1$

STEP 4

The asymptotes of the hyperbola are given by the equations $y = \pm \frac{a}{b}x$ .

STEP 5

Substitute the values of a and b into the equations for the asymptotes.
$y = \pm \frac{4}{1}x$ $y = \pm4x$

STEP 6

The equations of the asymptotes of the hyperbola are $y =4x$ and $y = -4x$ .

STEP 7

To graph the hyperbola, we need to plot the center, vertices, and asymptotes.The center is at the origin (0,0).The vertices are at (0,4) and (0,-4) because the hyperbola opens upwards and downwards as the y-term is positive in the equation.The asymptotes are the lines $y =4x$ and $y = -4x$ .

STEP 8

Draw the graph of the hyperbola using the information from the previous step.

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QuestionGraph the hyperbola $\frac{y^{2}}{16}-x^{2}=1$ and determine its asymptotes.

Studdy Solution

STEP 1

Assumptions1. The given equation is of a hyperbola. The equation is in the standard form $\frac{y^{}}{a^{}} - \frac{x^{}}{b^{}} =1$
3. The center of the hyperbola is at the origin (0,0)

STEP 2

From the given equation, we can identify the values of $a^{2}$ and $b^{2}$ . Here, $a^{2} =16$ and $b^{2} =1$ .

STEP 3

To find the values of a and b, we take the square root of $a^{2}$ and $b^{2}$ .
$a = \sqrt{16} =$ $b = \sqrt{1} =1$

STEP 4

The asymptotes of the hyperbola are given by the equations $y = \pm \frac{a}{b}x$ .

STEP 5

Substitute the values of a and b into the equations for the asymptotes.
$y = \pm \frac{4}{1}x$ $y = \pm4x$

STEP 6

The equations of the asymptotes of the hyperbola are $y =4x$ and $y = -4x$ .

STEP 7

To graph the hyperbola, we need to plot the center, vertices, and asymptotes.The center is at the origin (0,0).The vertices are at (0,4) and (0,-4) because the hyperbola opens upwards and downwards as the y-term is positive in the equation.The asymptotes are the lines $y =4x$ and $y = -4x$ .

STEP 8

Draw the graph of the hyperbola using the information from the previous step.

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QuestionGraph the hyperbola y216−x2=1\frac{y^{2}}{16}-x^{2}=116y2​−x2=1 and determine its asymptotes.

Studdy Solution

STEP 1

Assumptions1. The given equation is of a hyperbola. The equation is in the standard form ya−xb=1\frac{y^{}}{a^{}} - \frac{x^{}}{b^{}} =1ay​−bx​=13. The center of the hyperbola is at the origin (0,0)

STEP 2

From the given equation, we can identify the values of a2a^{2}a2 and b2b^{2}b2. Here, a2=16a^{2} =16a2=16 and b2=1b^{2} =1b2=1.

STEP 3

To find the values of a and b, we take the square root of a2a^{2}a2 and b2b^{2}b2.a=16=a = \sqrt{16} =a=16​=b=1=1b = \sqrt{1} =1b=1​=1

STEP 4

The asymptotes of the hyperbola are given by the equations y=±abxy = \pm \frac{a}{b}xy=±ba​x.

STEP 5

Substitute the values of a and b into the equations for the asymptotes.y=±41xy = \pm \frac{4}{1}xy=±14​xy=±4xy = \pm4xy=±4x

STEP 6

The equations of the asymptotes of the hyperbola are y=4xy =4xy=4x and y=−4xy = -4xy=−4x.

STEP 7

STEP 8

Draw the graph of the hyperbola using the information from the previous step.

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QuestionGraph the hyperbola $\frac{y^{2}}{16}-x^{2}=1$ and determine its asymptotes.

Assumptions1. The given equation is of a hyperbola. The equation is in the standard form $\frac{y^{}}{a^{}} - \frac{x^{}}{b^{}} =1$
3. The center of the hyperbola is at the origin (0,0)

From the given equation, we can identify the values of $a^{2}$ and $b^{2}$ . Here, $a^{2} =16$ and $b^{2} =1$ .

To find the values of a and b, we take the square root of $a^{2}$ and $b^{2}$ .
$a = \sqrt{16} =$ $b = \sqrt{1} =1$

The asymptotes of the hyperbola are given by the equations $y = \pm \frac{a}{b}x$ .

Substitute the values of a and b into the equations for the asymptotes.
$y = \pm \frac{4}{1}x$ $y = \pm4x$

The equations of the asymptotes of the hyperbola are $y =4x$ and $y = -4x$ .