Math / Algebra

Question $144y^2 - x^2 = 1$
Graph the hyperbola. Choose the correct graph below.
The foci is/are at the point(s) $\square$ . (Type an ordered pair. Type an exact answer, using radicals as needed. Use a comma to separate answers as needed.)
The equation of the asymptote with the positive slope is $\square$ . The equation of the asymptote with the negative slope is $\square$ . (Simplify your answers. Use integers or fractions for any numbers in the equation.)

Studdy Solution

STEP 1

What is this asking? We need to graph the hyperbola $144y^2 - x^2 = 1$ , find its foci, and the equations of its asymptotes. Watch out! The $y^2$ term comes first, so this hyperbola opens vertically, not horizontally!

STEP 2

1. Rewrite in standard form
2. Identify key features
3. Find the foci
4. Find the asymptotes
5. Choose the correct graph

STEP 3

We start with $144y^2 - x^2 = 1$ .
To get this into standard form, we need the right side to equal one.
It already does!
Woohoo! So we have $\frac{y^2}{\frac{1}{144}} - \frac{x^2}{1} = 1$ which is equivalent to $\frac{y^2}{(\frac{1}{12})^2} - \frac{x^2}{1^2} = 1$ .

STEP 4

This hyperbola is centered at the origin $(0,0)$ .
Since the $y^2$ term is positive, it opens vertically.
The value under $y^2$ gives us $a^2 = \frac{1}{144}$ , so $a = \frac{1}{12}$ .
The value under $x^2$ gives us $b^2 = 1$ , so $b = 1$ .

STEP 5

For hyperbolas, we use the equation $c^2 = a^2 + b^2$ to find the focal distance $c$ .
Plugging in our values, we get $c^2 = \frac{1}{144} + 1 = \frac{145}{144}$ , so $c = \frac{\sqrt{145}}{12}$ .

STEP 6

Since our hyperbola opens vertically, the foci are at $(0, \pm c)$ .
That means the foci are at $(0, \frac{\sqrt{145}}{12})$ and $(0, -\frac{\sqrt{145}}{12})$ .

STEP 7

The equations of the asymptotes for a vertical hyperbola centered at the origin are $y = \pm \frac{a}{b}x$ .
In our case, that's $y = \pm \frac{1/12}{1}x$ , which simplifies to $y = \frac{1}{12}x$ and $y = -\frac{1}{12}x$ .

STEP 8

Our hyperbola opens vertically, has vertices at $(0, \pm \frac{1}{12})$ , and asymptotes with slopes of $\pm \frac{1}{12}$ .
This matches graph B.

STEP 9

The foci are at $(0, \frac{\sqrt{145}}{12})$ and $(0, -\frac{\sqrt{145}}{12})$ .
The equation of the asymptote with positive slope is $y = \frac{1}{12}x$ .
The equation of the asymptote with negative slope is $y = -\frac{1}{12}x$ .
The correct graph is B.

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Studdy Solution

STEP 1

What is this asking? We need to graph the hyperbola $144y^2 - x^2 = 1$ , find its foci, and the equations of its asymptotes. Watch out! The $y^2$ term comes first, so this hyperbola opens vertically, not horizontally!

STEP 2

1. Rewrite in standard form
2. Identify key features
3. Find the foci
4. Find the asymptotes
5. Choose the correct graph

STEP 3

We start with $144y^2 - x^2 = 1$ .
To get this into standard form, we need the right side to equal one.
It already does!
Woohoo! So we have $\frac{y^2}{\frac{1}{144}} - \frac{x^2}{1} = 1$ which is equivalent to $\frac{y^2}{(\frac{1}{12})^2} - \frac{x^2}{1^2} = 1$ .

STEP 4

This hyperbola is centered at the origin $(0,0)$ .
Since the $y^2$ term is positive, it opens vertically.
The value under $y^2$ gives us $a^2 = \frac{1}{144}$ , so $a = \frac{1}{12}$ .
The value under $x^2$ gives us $b^2 = 1$ , so $b = 1$ .

STEP 5

For hyperbolas, we use the equation $c^2 = a^2 + b^2$ to find the focal distance $c$ .
Plugging in our values, we get $c^2 = \frac{1}{144} + 1 = \frac{145}{144}$ , so $c = \frac{\sqrt{145}}{12}$ .

STEP 6

Since our hyperbola opens vertically, the foci are at $(0, \pm c)$ .
That means the foci are at $(0, \frac{\sqrt{145}}{12})$ and $(0, -\frac{\sqrt{145}}{12})$ .

STEP 7

The equations of the asymptotes for a vertical hyperbola centered at the origin are $y = \pm \frac{a}{b}x$ .
In our case, that's $y = \pm \frac{1/12}{1}x$ , which simplifies to $y = \frac{1}{12}x$ and $y = -\frac{1}{12}x$ .

STEP 8

Our hyperbola opens vertically, has vertices at $(0, \pm \frac{1}{12})$ , and asymptotes with slopes of $\pm \frac{1}{12}$ .
This matches graph B.

STEP 9

The foci are at $(0, \frac{\sqrt{145}}{12})$ and $(0, -\frac{\sqrt{145}}{12})$ .
The equation of the asymptote with positive slope is $y = \frac{1}{12}x$ .
The equation of the asymptote with negative slope is $y = -\frac{1}{12}x$ .
The correct graph is B.

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Studdy Solution

STEP 1

What is this asking? We need to graph the hyperbola 144y2−x2=1144y^2 - x^2 = 1144y2−x2=1, find its foci, and the equations of its asymptotes. Watch out! The y2y^2y2 term comes first, so this hyperbola opens vertically, not horizontally!

STEP 2

1. Rewrite in standard form2. Identify key features3. Find the foci4. Find the asymptotes5. Choose the correct graph

STEP 3

STEP 4

This hyperbola is **centered at the origin** (0,0)(0,0)(0,0).Since the y2y^2y2 term is positive, it opens vertically.The value under y2y^2y2 gives us a2=1144a^2 = \frac{1}{144}a2=1441​, so a=112a = \frac{1}{12}a=121​.The value under x2x^2x2 gives us b2=1b^2 = 1b2=1, so b=1b = 1b=1.

STEP 5

For hyperbolas, we use the equation c2=a2+b2c^2 = a^2 + b^2c2=a2+b2 to find the **focal distance** ccc.Plugging in our values, we get c2=1144+1=145144c^2 = \frac{1}{144} + 1 = \frac{145}{144}c2=1441​+1=144145​, so c=14512c = \frac{\sqrt{145}}{12}c=12145​​.

STEP 6

Since our hyperbola opens vertically, the **foci** are at (0,±c)(0, \pm c)(0,±c).That means the foci are at (0,14512)(0, \frac{\sqrt{145}}{12})(0,12145​​) and (0,−14512)(0, -\frac{\sqrt{145}}{12})(0,−12145​​).

STEP 7

The **equations of the asymptotes** for a vertical hyperbola centered at the origin are y=±abxy = \pm \frac{a}{b}xy=±ba​x.In our case, that's y=±1/121xy = \pm \frac{1/12}{1}xy=±11/12​x, which simplifies to y=112xy = \frac{1}{12}xy=121​x and y=−112xy = -\frac{1}{12}xy=−121​x.

STEP 8

Our hyperbola opens vertically, has vertices at (0,±112)(0, \pm \frac{1}{12})(0,±121​), and asymptotes with slopes of ±112\pm \frac{1}{12}±121​.This matches **graph B**.

STEP 9

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What is this asking? We need to graph the hyperbola $144y^2 - x^2 = 1$ , find its foci, and the equations of its asymptotes. Watch out! The $y^2$ term comes first, so this hyperbola opens vertically, not horizontally!

1. Rewrite in standard form
2. Identify key features
3. Find the foci
4. Find the asymptotes
5. Choose the correct graph

This hyperbola is centered at the origin $(0,0)$ .
Since the $y^2$ term is positive, it opens vertically.
The value under $y^2$ gives us $a^2 = \frac{1}{144}$ , so $a = \frac{1}{12}$ .
The value under $x^2$ gives us $b^2 = 1$ , so $b = 1$ .

For hyperbolas, we use the equation $c^2 = a^2 + b^2$ to find the focal distance $c$ .
Plugging in our values, we get $c^2 = \frac{1}{144} + 1 = \frac{145}{144}$ , so $c = \frac{\sqrt{145}}{12}$ .

Since our hyperbola opens vertically, the foci are at $(0, \pm c)$ .
That means the foci are at $(0, \frac{\sqrt{145}}{12})$ and $(0, -\frac{\sqrt{145}}{12})$ .

The equations of the asymptotes for a vertical hyperbola centered at the origin are $y = \pm \frac{a}{b}x$ .
In our case, that's $y = \pm \frac{1/12}{1}x$ , which simplifies to $y = \frac{1}{12}x$ and $y = -\frac{1}{12}x$ .

Our hyperbola opens vertically, has vertices at $(0, \pm \frac{1}{12})$ , and asymptotes with slopes of $\pm \frac{1}{12}$ .
This matches graph B.