Recall that the standard formula of a hyperbola can come in two forms:

\(\displaystyle \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\) and

\(\displaystyle \frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}-=1\), where the centers of both hyperbolas are \(\displaystyle (h, k)\).

When the term with \(\displaystyle x\) is first, that means the foci will lie on a horizontal transverse axis.

When the term with \(\displaystyle y\) is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

\(\displaystyle c=\sqrt{a^2+b^2}\)

For a hyperbola with a horizontal transverse access, the foci will be located at \(\displaystyle (h+c, k)\) and \(\displaystyle (h-c, k)\).

For a hyperbola with a vertical transverse access, the foci will be located at \(\displaystyle (h, k+c)\) and \(\displaystyle (h, k-c)\).

For the given hypebola in the question, the transverse axis is horizontal and its center is located at \(\displaystyle (-2,2)\).

Next, find \(\displaystyle c\).

\(\displaystyle c=\sqrt{16+9}=\sqrt{25}=5\)

The foci are then located at \(\displaystyle (3, 2)\) and \(\displaystyle (-7, 2)\).

Recall that the standard formula of a hyperbola can come in two forms:

\(\displaystyle \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\) and

\(\displaystyle \frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}-=1\), where the centers of both hyperbolas are \(\displaystyle (h, k)\).

First, put the given equation in the standard form of the equation of a hyperbola.

Group the \(\displaystyle x\) terms together and the \(\displaystyle y\) terms together.

\(\displaystyle 50x^2-14y^2-400x+56y+44=0\)

\(\displaystyle 50x^2-400x-14y^2+56y+44=0\)

Next, factor out \(\displaystyle 50\) from the \(\displaystyle x\) terms and \(\displaystyle -14\) from the \(\displaystyle y\) terms.

\(\displaystyle 50(x^2-8x)-14(y^2-4y)+44=0\)

From here, complete the squares. Remember to add the same amount to both sides of the equation!

\(\displaystyle 50(x^2-8x+16)-14(y^2-4y+4)+44=744\)

Subtract \(\displaystyle 44\) from both sides.

\(\displaystyle 50(x^2-8x+16)-14(y^2-4y+4)=700\)

Divide both sides by \(\displaystyle 700\).

\(\displaystyle \frac{x^2-8x+16}{14}-\frac{y^2-4y+4}{50}=1\)

Factor both terms to get the standard form for the equation of a hyperbola.

\(\displaystyle \frac{(x-4)^2}{14}-\frac{(y-2)^2}{50}=1\)

When the term with \(\displaystyle x\) is first, that means the foci will lie on a horizontal transverse axis.

When the term with \(\displaystyle y\) is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

\(\displaystyle c=\sqrt{a^2+b^2}\)

For a hyperbola with a horizontal transverse access, the foci will be located at \(\displaystyle (h+c, k)\) and \(\displaystyle (h-c, k)\).

For a hyperbola with a vertical transverse access, the foci will be located at \(\displaystyle (h, k+c)\) and \(\displaystyle (h, k-c)\).

For the given hypebola in the question, the transverse axis is horizontal and its center is located at \(\displaystyle (4,2)\).

Next, find \(\displaystyle c\).

\(\displaystyle c=\sqrt{14+50}=\sqrt{64}=8\)

The foci are then located at \(\displaystyle (12, 2)\) and \(\displaystyle (-4, 2)\).

Recall that the standard formula of a hyperbola can come in two forms:

\(\displaystyle \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\) and

\(\displaystyle \frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}-=1\), where the centers of both hyperbolas are \(\displaystyle (h, k)\).

First, put the given equation in the standard form of the equation of a hyperbola.

Group the \(\displaystyle x\) terms together and the \(\displaystyle y\) terms together.

\(\displaystyle 12x^2-4y^2-216x-8y+920=0\)

\(\displaystyle 12x^2-216x-4y^2-8y+920=0\)

Next, factor out \(\displaystyle 12\) from the \(\displaystyle x\) terms and \(\displaystyle -4\) from the \(\displaystyle y\) terms.

\(\displaystyle 12(x^2-18x)-4(y^2+2y)+920=0\)

From here, complete the squares. Remember to add the same amount to both sides of the equation!

\(\displaystyle 12(x^2-18x+81)-4(y^2+2y+1)+920=968\)

Subtract \(\displaystyle 920\) from both sides.

\(\displaystyle 12(x^2-18x+81)-4(y^2+2y+1)=48\)

Divide both sides by \(\displaystyle 48\).

\(\displaystyle \frac{x^2-18x+81}{4}-\frac{y^2+2y+1}{12}=1\)

Factor both terms to get the standard form for the equation of a hyperbola.

\(\displaystyle \frac{(x-9)^2}{4}-\frac{(y+1)^2}{12}=1\)

When the term with \(\displaystyle x\) is first, that means the foci will lie on a horizontal transverse axis.

When the term with \(\displaystyle y\) is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

\(\displaystyle c=\sqrt{a^2+b^2}\)

For a hyperbola with a horizontal transverse access, the foci will be located at \(\displaystyle (h+c, k)\) and \(\displaystyle (h-c, k)\).

For a hyperbola with a vertical transverse access, the foci will be located at \(\displaystyle (h, k+c)\) and \(\displaystyle (h, k-c)\).

For the given hypebola in the question, the transverse axis is horizontal and its center is located at \(\displaystyle (9, -1)\).

Next, find \(\displaystyle c\).

\(\displaystyle c=\sqrt{12+4}=\sqrt{16}=4\)

The foci are then located at \(\displaystyle (13, -1)\) and \(\displaystyle (5, -1)\).

Recall that the standard formula of a hyperbola can come in two forms:

\(\displaystyle \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\) and

\(\displaystyle \frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}-=1\), where the centers of both hyperbolas are \(\displaystyle (h, k)\).

First, put the given equation in the standard form of the equation of a hyperbola.

Group the \(\displaystyle x\) terms together and the \(\displaystyle y\) terms together.

\(\displaystyle 20y^2-16x^2-120y+64x-204=0\)

\(\displaystyle 20y^2-120y-16x^2+64x-204=0\)

Next, factor out \(\displaystyle -16\) from the \(\displaystyle x\) terms and \(\displaystyle 20\) from the \(\displaystyle y\) terms.

\(\displaystyle 20(y^2-6y)-4(x^2-4x)-204=0\)

Complete the squares. Remember to add the same amount to both sides of the equation!

\(\displaystyle 20(y^2-6y+9)-4(x^2-4x+4)-204=116\)

Add \(\displaystyle 204\) to both sides.

\(\displaystyle 20(y^2-6y+9)-4(x^2-4x+4)=320\)

Divide both sides by \(\displaystyle 320\).

\(\displaystyle \frac{y^2-6y+9}{16}-\frac{x^2-4x+4}{20}=1\)

Factor both terms to get the standard form of the equation of a hyperbola.

\(\displaystyle \frac{(y-3)^2}{16}-\frac{(x-2)^2}{20}=1\)

When the term with \(\displaystyle x\) is first, that means the foci will lie on a horizontal transverse axis.

When the term with \(\displaystyle y\) is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

\(\displaystyle c=\sqrt{a^2+b^2}\)

For a hyperbola with a horizontal transverse access, the foci will be located at \(\displaystyle (h+c, k)\) and \(\displaystyle (h-c, k)\).

For a hyperbola with a vertical transverse access, the foci will be located at \(\displaystyle (h, k+c)\) and \(\displaystyle (h, k-c)\).

For the given hypebola in the question, the transverse axis is vertical and its center is located at \(\displaystyle (2,3)\).

Next, find \(\displaystyle c\).

\(\displaystyle c=\sqrt{16+20}=\sqrt{36}=6\)

The foci are then located at \(\displaystyle (2, 9)\) and \(\displaystyle (2, -3)\).

Recall that the standard formula of a hyperbola can come in two forms:

\(\displaystyle \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\) and

\(\displaystyle \frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}-=1\), where the centers of both hyperbolas are \(\displaystyle (h, k)\).

First, put the given equation in the standard form of the equation of a hyperbola.

Group the \(\displaystyle x\) terms together and the \(\displaystyle y\) terms together.

\(\displaystyle 22y^2-27x^2+220y-54x-71=0\)

\(\displaystyle 22y^2+220y-27x^2-54x-71=0\)

Next, factor out \(\displaystyle -27\) from the \(\displaystyle x\) terms and \(\displaystyle 22\) from the \(\displaystyle y\) terms.

\(\displaystyle 22(y^2+10y)-27(x^2+2x)-71=0\)

Now we can complete the squares. Remember to add the same amount to both sides of the equation!

\(\displaystyle 22(y^2+10y+25)-27(x^2+2x+1)-71=523\)

Add \(\displaystyle 71\) to both sides.

\(\displaystyle 22(y^2+10y+25)-27(x^2+2x+1)-71=594\)

Divide both sides by \(\displaystyle 594\).

\(\displaystyle \frac{y^2+10y+25}{27}-\frac{x^2+2x+1}{22}=1\)

Factor both terms to get the standard form of the equation of a hyperbola.

\(\displaystyle \frac{(y+5)^2}{27}-\frac{(x+1)^2}{22}=1\)

When the term with \(\displaystyle x\) is first, that means the foci will lie on a horizontal transverse axis.

When the term with \(\displaystyle y\) is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

\(\displaystyle c=\sqrt{a^2+b^2}\)

For a hyperbola with a horizontal transverse access, the foci will be located at \(\displaystyle (h+c, k)\) and \(\displaystyle (h-c, k)\).

For a hyperbola with a vertical transverse access, the foci will be located at \(\displaystyle (h, k+c)\) and \(\displaystyle (h, k-c)\).

For the given hypebola in the question, the transverse axis is vertical and its center is located at \(\displaystyle (-1,-5)\).

Next, find \(\displaystyle c\).

\(\displaystyle c=\sqrt{22+27}=\sqrt{49}=7\)

The foci are then located at \(\displaystyle (-1, 2)\) and \(\displaystyle (-1, -12)\).

Recall that the standard formula of a hyperbola can come in two forms:

\(\displaystyle \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\) and

\(\displaystyle \frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}-=1\), where the centers of both hyperbolas are \(\displaystyle (h, k)\).

When the term with \(\displaystyle x\) is first, that means the foci will lie on a horizontal transverse axis.

When the term with \(\displaystyle y\) is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

\(\displaystyle c=\sqrt{a^2+b^2}\)

For a hyperbola with a horizontal transverse access, the foci will be located at \(\displaystyle (h+c, k)\) and \(\displaystyle (h-c, k)\).

For a hyperbola with a vertical transverse access, the foci will be located at \(\displaystyle (h, k+c)\) and \(\displaystyle (h, k-c)\).

For the given hypebola in the question, the transverse axis is vertical and its center is located at \(\displaystyle (-6,7)\).

Next, find \(\displaystyle c\).

\(\displaystyle c=\sqrt{9+7}=\sqrt{16}=4\)

The foci are then located at \(\displaystyle (-6,11)\) and \(\displaystyle (-6,3)\).

Recall that the standard formula of a hyperbola can come in two forms:

\(\displaystyle \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\) and

\(\displaystyle \frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}-=1\), where the centers of both hyperbolas are \(\displaystyle (h, k)\).

When the term with \(\displaystyle x\) is first, that means the foci will lie on a horizontal transverse axis.

When the term with \(\displaystyle y\) is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

\(\displaystyle c=\sqrt{a^2+b^2}\)

For a hyperbola with a horizontal transverse access, the foci will be located at \(\displaystyle (h+c, k)\) and \(\displaystyle (h-c, k)\).

For a hyperbola with a vertical transverse access, the foci will be located at \(\displaystyle (h, k+c)\) and \(\displaystyle (h, k-c)\).

For the given hypebola in the question, the transverse axis is vertical and its center is located at \(\displaystyle (5,1)\).

Next, find \(\displaystyle c\).

\(\displaystyle c=\sqrt{14+22}=\sqrt{36}=6\)

The foci are then located at \(\displaystyle (5, 7)\) and \(\displaystyle (5, -5)\).

Recall that the standard formula of a hyperbola can come in two forms:

\(\displaystyle \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\) and

\(\displaystyle \frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}-=1\), where the centers of both hyperbolas are \(\displaystyle (h, k)\).

When the term with \(\displaystyle x\) is first, that means the foci will lie on a horizontal transverse axis.

When the term with \(\displaystyle y\) is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

\(\displaystyle c=\sqrt{a^2+b^2}\)

For a hyperbola with a horizontal transverse access, the foci will be located at \(\displaystyle (h+c, k)\) and \(\displaystyle (h-c, k)\).

For a hyperbola with a vertical transverse access, the foci will be located at \(\displaystyle (h, k+c)\) and \(\displaystyle (h, k-c)\).

For the given hypebola in the question, the transverse axis is vertical and its center is located at \(\displaystyle (0,0)\).

Next, find \(\displaystyle c\).

\(\displaystyle c=\sqrt{14+2}=\sqrt{16}=4\)

The foci are then located at \(\displaystyle (0, 4)\) and \(\displaystyle (0, -4)\).

Recall that the standard formula of a hyperbola can come in two forms:

\(\displaystyle \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\) and

\(\displaystyle \frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}-=1\), where the centers of both hyperbolas are \(\displaystyle (h, k)\).

When the term with \(\displaystyle x\) is first, that means the foci will lie on a horizontal transverse axis.

When the term with \(\displaystyle y\) is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

\(\displaystyle c=\sqrt{a^2+b^2}\)

For a hyperbola with a horizontal transverse access, the foci will be located at \(\displaystyle (h+c, k)\) and \(\displaystyle (h-c, k)\).

For a hyperbola with a vertical transverse access, the foci will be located at \(\displaystyle (h, k+c)\) and \(\displaystyle (h, k-c)\).

For the given hypebola in the question, the transverse axis is vertical and its center is located at \(\displaystyle (0,0)\).

Next, find \(\displaystyle c\).

\(\displaystyle c=\sqrt{12+13}=\sqrt{25}=5\)

The foci are then located at \(\displaystyle (0, 5)\) and \(\displaystyle (0, -5)\).

To find the foci of a hyperbola, first determine a and b, and then use the relationship \(\displaystyle b^2 = c^2 - a^2\)

In this case, the major axis is horizontal since x comes first, so \(\displaystyle a^2 = 9\) and \(\displaystyle b^2 = 2\).

Solve for c: \(\displaystyle 2 = c^2 - 9\) add 9 to both sides 

\(\displaystyle 11 = c^2\) take the square root

\(\displaystyle \sqrt{11} = c\)

Since the center is \(\displaystyle (5, -1)\) and the major axis is the horizontal one, our foci are \(\displaystyle (5 \pm \sqrt{11} , -1)\). The only choice that works is

\(\displaystyle (5 + \sqrt{11}, -1 )\).