Math / Geometry

QuestionFind the foci of the hyperbola defined by the equation $\frac{(y+4)^{2}}{81}-\frac{(x+8)^{2}}{49}=1$ . If necessary round to the nearest tenth.
Answer Attempt 1 out of 2
Foci: $\square$ , $\square$ ) and ( $\square$ , , )

Studdy Solution

STEP 1

What is this asking? We need to find the two focal points of a hyperbola centered at $(-8, -4)$ with a vertical main axis. Watch out! Don't mix up the formulas for ellipses and hyperbolas!
Also, remember the center isn't at $(0, 0)$ !

STEP 2

1. Identify key features of the hyperbola.
2. Calculate $c$ .
3. Find the foci.

STEP 3

Alright, let's break down this hyperbola equation!
We've got $\frac{(y+4)^{2}}{81}-\frac{(x+8)^{2}}{49}=1$ .
This tells us a few things.
First, the center of the hyperbola is at $(-8, -4)$ .

STEP 4

Next, since the $y$ term is positive, we know this hyperbola opens vertically.
The value under the $y$ term is $81 = a^2$ , so $a = 9$ .
The value under the $x$ term is $49 = b^2$ , so $b = 7$ .

STEP 5

For hyperbolas, the relationship between $a$ , $b$ , and $c$ (where $c$ is the distance from the center to each focus) is $c^2 = a^2 + b^2$ .
It's like the Pythagorean theorem, but with a plus instead of a minus!

STEP 6

Let's plug in our values: $c^2 = 9^2 + 7^2 = 81 + 49 = 130$ .
So, $c = \sqrt{130} \approx \mathbf{11.4}$ .

STEP 7

Since our hyperbola opens vertically, the foci will be $c$ units above and below the center.
The x-coordinate of the foci will be the same as the center's x-coordinate, which is $-8$ .

STEP 8

The y-coordinates of the foci will be the center's y-coordinate, $-4$ , plus or minus $c$ .
So, the foci are at $(-8, -4 + \sqrt{130})$ and $(-8, -4 - \sqrt{130})$ .

STEP 9

Using our approximation for $c$ , the foci are approximately at $(-8, -4 + 11.4)$ and $(-8, -4 - 11.4)$ , which simplifies to $(-8, 7.4)$ and $(-8, -15.4)$ .

STEP 10

Foci: $(-8, 7.4)$ and $(-8, -15.4)$

Was this helpful?

QuestionFind the foci of the hyperbola defined by the equation $\frac{(y+4)^{2}}{81}-\frac{(x+8)^{2}}{49}=1$ . If necessary round to the nearest tenth.
Answer Attempt 1 out of 2
Foci: $\square$ , $\square$ ) and ( $\square$ , , )

Studdy Solution

STEP 1

What is this asking? We need to find the two focal points of a hyperbola centered at $(-8, -4)$ with a vertical main axis. Watch out! Don't mix up the formulas for ellipses and hyperbolas!
Also, remember the center isn't at $(0, 0)$ !

STEP 2

1. Identify key features of the hyperbola.
2. Calculate $c$ .
3. Find the foci.

STEP 3

Alright, let's break down this hyperbola equation!
We've got $\frac{(y+4)^{2}}{81}-\frac{(x+8)^{2}}{49}=1$ .
This tells us a few things.
First, the center of the hyperbola is at $(-8, -4)$ .

STEP 4

Next, since the $y$ term is positive, we know this hyperbola opens vertically.
The value under the $y$ term is $81 = a^2$ , so $a = 9$ .
The value under the $x$ term is $49 = b^2$ , so $b = 7$ .

STEP 5

For hyperbolas, the relationship between $a$ , $b$ , and $c$ (where $c$ is the distance from the center to each focus) is $c^2 = a^2 + b^2$ .
It's like the Pythagorean theorem, but with a plus instead of a minus!

STEP 6

Let's plug in our values: $c^2 = 9^2 + 7^2 = 81 + 49 = 130$ .
So, $c = \sqrt{130} \approx \mathbf{11.4}$ .

STEP 7

Since our hyperbola opens vertically, the foci will be $c$ units above and below the center.
The x-coordinate of the foci will be the same as the center's x-coordinate, which is $-8$ .

STEP 8

The y-coordinates of the foci will be the center's y-coordinate, $-4$ , plus or minus $c$ .
So, the foci are at $(-8, -4 + \sqrt{130})$ and $(-8, -4 - \sqrt{130})$ .

STEP 9

Using our approximation for $c$ , the foci are approximately at $(-8, -4 + 11.4)$ and $(-8, -4 - 11.4)$ , which simplifies to $(-8, 7.4)$ and $(-8, -15.4)$ .

STEP 10

Foci: $(-8, 7.4)$ and $(-8, -15.4)$

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionFind the foci of the hyperbola defined by the equation (y+4)281−(x+8)249=1\frac{(y+4)^{2}}{81}-\frac{(x+8)^{2}}{49}=181(y+4)2​−49(x+8)2​=1. If necessary round to the nearest tenth.Answer Attempt 1 out of 2Foci: □\square□ , □\square□ ) and ( □\square□ , , )

Studdy Solution

STEP 1

What is this asking? We need to find the two focal points of a hyperbola centered at (−8,−4)(-8, -4)(−8,−4) with a vertical main axis. Watch out! Don't mix up the formulas for ellipses and hyperbolas!Also, remember the center isn't at (0,0)(0, 0)(0,0)!

STEP 2

1. Identify key features of the hyperbola.2. Calculate ccc.3. Find the foci.

STEP 3

Alright, let's **break down** this hyperbola equation!We've got (y+4)281−(x+8)249=1\frac{(y+4)^{2}}{81}-\frac{(x+8)^{2}}{49}=181(y+4)2​−49(x+8)2​=1.This tells us a few things.First, the **center** of the hyperbola is at (−8,−4)(-8, -4)(−8,−4).

STEP 4

Next, since the yyy term is positive, we know this hyperbola opens **vertically**.The value under the yyy term is 81=a281 = a^281=a2, so a=9a = 9a=9.The value under the xxx term is 49=b249 = b^249=b2, so b=7b = 7b=7.

STEP 5

For hyperbolas, the relationship between aaa, bbb, and ccc (where ccc is the distance from the center to each focus) is c2=a2+b2c^2 = a^2 + b^2c2=a2+b2.It's like the Pythagorean theorem, but with a **plus** instead of a minus!

STEP 6

Let's **plug in** our values: c2=92+72=81+49=130c^2 = 9^2 + 7^2 = 81 + 49 = 130c2=92+72=81+49=130.So, c=130≈11.4c = \sqrt{130} \approx \mathbf{11.4}c=130​≈11.4.

STEP 7

Since our hyperbola opens vertically, the foci will be ccc units **above and below** the center.The x-coordinate of the foci will be the same as the center's x-coordinate, which is −8-8−8.

STEP 8

The y-coordinates of the foci will be the center's y-coordinate, −4-4−4, plus or minus ccc.So, the foci are at (−8,−4+130)(-8, -4 + \sqrt{130})(−8,−4+130​) and (−8,−4−130)(-8, -4 - \sqrt{130})(−8,−4−130​).

STEP 9

Using our approximation for ccc, the foci are approximately at (−8,−4+11.4)(-8, -4 + 11.4)(−8,−4+11.4) and (−8,−4−11.4)(-8, -4 - 11.4)(−8,−4−11.4), which simplifies to (−8,7.4)(-8, 7.4)(−8,7.4) and (−8,−15.4)(-8, -15.4)(−8,−15.4).

STEP 10

Foci: (−8,7.4)(-8, 7.4)(−8,7.4) and (−8,−15.4)(-8, -15.4)(−8,−15.4)

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionFind the foci of the hyperbola defined by the equation $\frac{(y+4)^{2}}{81}-\frac{(x+8)^{2}}{49}=1$ . If necessary round to the nearest tenth.
Answer Attempt 1 out of 2
Foci: $\square$ , $\square$ ) and ( $\square$ , , )

What is this asking? We need to find the two focal points of a hyperbola centered at $(-8, -4)$ with a vertical main axis. Watch out! Don't mix up the formulas for ellipses and hyperbolas!
Also, remember the center isn't at $(0, 0)$ !

1. Identify key features of the hyperbola.
2. Calculate $c$ .
3. Find the foci.

Alright, let's break down this hyperbola equation!
We've got $\frac{(y+4)^{2}}{81}-\frac{(x+8)^{2}}{49}=1$ .
This tells us a few things.
First, the center of the hyperbola is at $(-8, -4)$ .

Next, since the $y$ term is positive, we know this hyperbola opens vertically.
The value under the $y$ term is $81 = a^2$ , so $a = 9$ .
The value under the $x$ term is $49 = b^2$ , so $b = 7$ .

For hyperbolas, the relationship between $a$ , $b$ , and $c$ (where $c$ is the distance from the center to each focus) is $c^2 = a^2 + b^2$ .
It's like the Pythagorean theorem, but with a plus instead of a minus!

Let's plug in our values: $c^2 = 9^2 + 7^2 = 81 + 49 = 130$ .
So, $c = \sqrt{130} \approx \mathbf{11.4}$ .

Since our hyperbola opens vertically, the foci will be $c$ units above and below the center.
The x-coordinate of the foci will be the same as the center's x-coordinate, which is $-8$ .

The y-coordinates of the foci will be the center's y-coordinate, $-4$ , plus or minus $c$ .
So, the foci are at $(-8, -4 + \sqrt{130})$ and $(-8, -4 - \sqrt{130})$ .

Using our approximation for $c$ , the foci are approximately at $(-8, -4 + 11.4)$ and $(-8, -4 - 11.4)$ , which simplifies to $(-8, 7.4)$ and $(-8, -15.4)$ .

Foci: $(-8, 7.4)$ and $(-8, -15.4)$