QuestionFind the vertices and foci of the hyperbola $\frac{y^{2}}{49}-\frac{x^{2}}{36}=1$ . Enter as $(0, \pm a)$ and $(0, \pm c)$ .

Studdy Solution

STEP 1

Assumptions1. The given equation is of a hyperbola. The hyperbola 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

The general form of the equation of a hyperbola centered at the origin with its transverse axis along the y-axis is given by $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$ . The vertices are given by $(0, \pm a)$ and the foci by $(0, \pm c)$ , where $c=\sqrt{a^{2}+b^{2}}$ .

STEP 3

From the equation $\frac{y^{2}}{49}-\frac{x^{2}}{36}=1$ , we can see that $a^{2}=49$ and $b^{2}=36$ .

STEP 4

To find the values of $a$ and $b$ , we take the square root of $a^{2}$ and $b^{2}$ .
$a=\sqrt{49}$ $b=\sqrt{36}$

STEP 5

Calculate the values of $a$ and $b$ .
$a=\sqrt{49}=7$ $b=\sqrt{36}=$

STEP 6

Now, we can find the vertices of the hyperbola. The vertices are given by $(0, \pm a)$ .
$Vertices = (0, \pm)$

STEP 7

Next, we need to find the foci of the hyperbola. The foci are given by $(0, \pm c)$ , where $c=\sqrt{a^{2}+b^{2}}$ .
$c=\sqrt{a^{2}+b^{2}}$

STEP 8

Plug in the values for $a$ and $b$ to calculate $c$ .
$c=\sqrt{7^{2}+6^{2}}$

STEP 9

Calculate the value of $c$ .
$c=\sqrt{49+36}=\sqrt{85}$

STEP 10

Now, we can find the foci of the hyperbola. The foci are given by $(0, \pm c)$ .
$oci = (0, \pm \sqrt{85})$ The vertices are $(0, \pm7)$ and the foci are $(0, \pm \sqrt{85})$ .

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QuestionFind the vertices and foci of the hyperbola $\frac{y^{2}}{49}-\frac{x^{2}}{36}=1$ . Enter as $(0, \pm a)$ and $(0, \pm c)$ .

Studdy Solution

STEP 1

Assumptions1. The given equation is of a hyperbola. The hyperbola 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

The general form of the equation of a hyperbola centered at the origin with its transverse axis along the y-axis is given by $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$ . The vertices are given by $(0, \pm a)$ and the foci by $(0, \pm c)$ , where $c=\sqrt{a^{2}+b^{2}}$ .

STEP 3

From the equation $\frac{y^{2}}{49}-\frac{x^{2}}{36}=1$ , we can see that $a^{2}=49$ and $b^{2}=36$ .

STEP 4

To find the values of $a$ and $b$ , we take the square root of $a^{2}$ and $b^{2}$ .
$a=\sqrt{49}$ $b=\sqrt{36}$

STEP 5

Calculate the values of $a$ and $b$ .
$a=\sqrt{49}=7$ $b=\sqrt{36}=$

STEP 6

Now, we can find the vertices of the hyperbola. The vertices are given by $(0, \pm a)$ .
$Vertices = (0, \pm)$

STEP 7

Next, we need to find the foci of the hyperbola. The foci are given by $(0, \pm c)$ , where $c=\sqrt{a^{2}+b^{2}}$ .
$c=\sqrt{a^{2}+b^{2}}$

STEP 8

Plug in the values for $a$ and $b$ to calculate $c$ .
$c=\sqrt{7^{2}+6^{2}}$

STEP 9

Calculate the value of $c$ .
$c=\sqrt{49+36}=\sqrt{85}$

STEP 10

Now, we can find the foci of the hyperbola. The foci are given by $(0, \pm c)$ .
$oci = (0, \pm \sqrt{85})$ The vertices are $(0, \pm7)$ and the foci are $(0, \pm \sqrt{85})$ .

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QuestionFind the vertices and foci of the hyperbola y249−x236=1\frac{y^{2}}{49}-\frac{x^{2}}{36}=149y2​−36x2​=1. Enter as (0,±a)(0, \pm a)(0,±a) and (0,±c)(0, \pm c)(0,±c).

Studdy Solution

STEP 1

Assumptions1. The given equation is of a hyperbola. The hyperbola 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

STEP 3

From the equation y249−x236=1\frac{y^{2}}{49}-\frac{x^{2}}{36}=149y2​−36x2​=1, we can see that a2=49a^{2}=49a2=49 and b2=36b^{2}=36b2=36.

STEP 4

To find the values of aaa and bbb, we take the square root of a2a^{2}a2 and b2b^{2}b2.a=49a=\sqrt{49}a=49​b=36b=\sqrt{36}b=36​

STEP 5

Calculate the values of aaa and bbb.a=49=7a=\sqrt{49}=7a=49​=7b=36=b=\sqrt{36}=b=36​=

STEP 6

Now, we can find the vertices of the hyperbola. The vertices are given by (0,±a)(0, \pm a)(0,±a).Vertices=(0,±)Vertices = (0, \pm)Vertices=(0,±)

STEP 7

Next, we need to find the foci of the hyperbola. The foci are given by (0,±c)(0, \pm c)(0,±c), where c=a2+b2c=\sqrt{a^{2}+b^{2}}c=a2+b2​.c=a2+b2c=\sqrt{a^{2}+b^{2}}c=a2+b2​

STEP 8

Plug in the values for aaa and bbb to calculate ccc.c=72+62c=\sqrt{7^{2}+6^{2}}c=72+62​

STEP 9

Calculate the value of ccc.c=49+36=85c=\sqrt{49+36}=\sqrt{85}c=49+36​=85​

STEP 10

Now, we can find the foci of the hyperbola. The foci are given by (0,±c)(0, \pm c)(0,±c).oci=(0,±85)oci = (0, \pm \sqrt{85})oci=(0,±85​)The vertices are (0,±7)(0, \pm7)(0,±7) and the foci are (0,±85)(0, \pm \sqrt{85})(0,±85​).

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QuestionFind the vertices and foci of the hyperbola $\frac{y^{2}}{49}-\frac{x^{2}}{36}=1$ . Enter as $(0, \pm a)$ and $(0, \pm c)$ .

Assumptions1. The given equation is of a hyperbola. The hyperbola 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 equation $\frac{y^{2}}{49}-\frac{x^{2}}{36}=1$ , we can see that $a^{2}=49$ and $b^{2}=36$ .

To find the values of $a$ and $b$ , we take the square root of $a^{2}$ and $b^{2}$ .
$a=\sqrt{49}$ $b=\sqrt{36}$

Calculate the values of $a$ and $b$ .
$a=\sqrt{49}=7$ $b=\sqrt{36}=$

Now, we can find the vertices of the hyperbola. The vertices are given by $(0, \pm a)$ .
$Vertices = (0, \pm)$

Next, we need to find the foci of the hyperbola. The foci are given by $(0, \pm c)$ , where $c=\sqrt{a^{2}+b^{2}}$ .
$c=\sqrt{a^{2}+b^{2}}$

Plug in the values for $a$ and $b$ to calculate $c$ .
$c=\sqrt{7^{2}+6^{2}}$

Calculate the value of $c$ .
$c=\sqrt{49+36}=\sqrt{85}$

Now, we can find the foci of the hyperbola. The foci are given by $(0, \pm c)$ .
$oci = (0, \pm \sqrt{85})$ The vertices are $(0, \pm7)$ and the foci are $(0, \pm \sqrt{85})$ .