Math / Geometry

QuestionFind the center, vertices, foci, and the equations of the asymptotes of the hyperbola. (If an answer does not exist, enter DNE.) $\frac{(x+4)^{2}}{144}-\frac{(y-2)^{2}}{25}=1$

Studdy Solution

STEP 1

1. The equation given is a standard form of a hyperbola.
2. The hyperbola is centered at a point $(h, k)$ .
3. The hyperbola opens horizontally since the $x$ -term is positive.

STEP 2

1. Identify the center of the hyperbola.
2. Determine the vertices of the hyperbola.
3. Find the foci of the hyperbola.
4. Derive the equations of the asymptotes.

STEP 3

Identify the center of the hyperbola:
The standard form of a hyperbola is:
$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$
Comparing with the given equation:
$\frac{(x+4)^2}{144} - \frac{(y-2)^2}{25} = 1$
The center $(h, k)$ is $(-4, 2)$ .

STEP 4

Determine the vertices of the hyperbola:
For a hyperbola that opens horizontally, the vertices are at $(h \pm a, k)$ .
Here, $a^2 = 144$ , so $a = 12$ .
Vertices: $(-4 \pm 12, 2)$ which are $(-16, 2)$ and $(8, 2)$ .

STEP 5

Find the foci of the hyperbola:
The distance to the foci from the center is given by $c$ , where $c^2 = a^2 + b^2$ .
Here, $b^2 = 25$ , so $b = 5$ .
Calculate $c$ :
$c^2 = 144 + 25 = 169 \implies c = 13$
Foci: $(-4 \pm 13, 2)$ which are $(-17, 2)$ and $(9, 2)$ .

STEP 6

Derive the equations of the asymptotes:
For a hyperbola that opens horizontally, the equations of the asymptotes are:
$y = k \pm \frac{b}{a}(x-h)$
Substitute the known values:
$y = 2 \pm \frac{5}{12}(x + 4)$
Simplify the equations:
1. $y = 2 + \frac{5}{12}(x + 4)$
2. $y = 2 - \frac{5}{12}(x + 4)$
These are the equations of the asymptotes.
The center is $(-4, 2)$ , the vertices are $(-16, 2)$ and $(8, 2)$ , the foci are $(-17, 2)$ and $(9, 2)$ , and the equations of the asymptotes are $y = 2 + \frac{5}{12}(x + 4)$ and $y = 2 - \frac{5}{12}(x + 4)$ .

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QuestionFind the center, vertices, foci, and the equations of the asymptotes of the hyperbola. (If an answer does not exist, enter DNE.) $\frac{(x+4)^{2}}{144}-\frac{(y-2)^{2}}{25}=1$

Studdy Solution

STEP 1

1. The equation given is a standard form of a hyperbola.
2. The hyperbola is centered at a point $(h, k)$ .
3. The hyperbola opens horizontally since the $x$ -term is positive.

STEP 2

1. Identify the center of the hyperbola.
2. Determine the vertices of the hyperbola.
3. Find the foci of the hyperbola.
4. Derive the equations of the asymptotes.

STEP 3

Identify the center of the hyperbola:
The standard form of a hyperbola is:
$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$
Comparing with the given equation:
$\frac{(x+4)^2}{144} - \frac{(y-2)^2}{25} = 1$
The center $(h, k)$ is $(-4, 2)$ .

STEP 4

Determine the vertices of the hyperbola:
For a hyperbola that opens horizontally, the vertices are at $(h \pm a, k)$ .
Here, $a^2 = 144$ , so $a = 12$ .
Vertices: $(-4 \pm 12, 2)$ which are $(-16, 2)$ and $(8, 2)$ .

STEP 5

Find the foci of the hyperbola:
The distance to the foci from the center is given by $c$ , where $c^2 = a^2 + b^2$ .
Here, $b^2 = 25$ , so $b = 5$ .
Calculate $c$ :
$c^2 = 144 + 25 = 169 \implies c = 13$
Foci: $(-4 \pm 13, 2)$ which are $(-17, 2)$ and $(9, 2)$ .

STEP 6

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QuestionFind the center, vertices, foci, and the equations of the asymptotes of the hyperbola. (If an answer does not exist, enter DNE.) (x+4)2144−(y−2)225=1\frac{(x+4)^{2}}{144}-\frac{(y-2)^{2}}{25}=1144(x+4)2​−25(y−2)2​=1

Studdy Solution

STEP 1

1. The equation given is a standard form of a hyperbola.2. The hyperbola is centered at a point (h,k)(h, k)(h,k).3. The hyperbola opens horizontally since the xxx-term is positive.

STEP 2

1. Identify the center of the hyperbola.2. Determine the vertices of the hyperbola.3. Find the foci of the hyperbola.4. Derive the equations of the asymptotes.

STEP 3

STEP 4

Determine the vertices of the hyperbola:For a hyperbola that opens horizontally, the vertices are at (h±a,k)(h \pm a, k)(h±a,k).Here, a2=144a^2 = 144a2=144, so a=12a = 12a=12.Vertices: (−4±12,2)(-4 \pm 12, 2)(−4±12,2) which are (−16,2)(-16, 2)(−16,2) and (8,2)(8, 2)(8,2).

STEP 5

STEP 6

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QuestionFind the center, vertices, foci, and the equations of the asymptotes of the hyperbola. (If an answer does not exist, enter DNE.) $\frac{(x+4)^{2}}{144}-\frac{(y-2)^{2}}{25}=1$

1. The equation given is a standard form of a hyperbola.
2. The hyperbola is centered at a point $(h, k)$ .
3. The hyperbola opens horizontally since the $x$ -term is positive.

1. Identify the center of the hyperbola.
2. Determine the vertices of the hyperbola.
3. Find the foci of the hyperbola.
4. Derive the equations of the asymptotes.

Determine the vertices of the hyperbola:
For a hyperbola that opens horizontally, the vertices are at $(h \pm a, k)$ .
Here, $a^2 = 144$ , so $a = 12$ .
Vertices: $(-4 \pm 12, 2)$ which are $(-16, 2)$ and $(8, 2)$ .