Hyperbolas - College Algebra

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Question

Using the information below, determine the equation of the hyperbola.

Foci: and

Eccentricity: $\frac{3}{2}$

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Answer

General Information for Hyperbola:

Equation for horizontal transverse hyperbola:

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

Distance between foci =

Distance between vertices =

Eccentricity =

$\small $a^{2}$$+b^{2}$$=c^{2}$$

Center: (h, k)

First determine the value of c. Since we know the distance between the two foci is 12, we can set that equal to .

$\small c= 6$

Next, use the eccentricity equation and the value of the eccentricity provided in the question to determine the value of a.

Eccentricity = $\small \small c/a=3/2$

$\small $\frac{c}{a}$=\frac{3}{2}$=\frac{6}{a}$$

$\small 12=3a$

$\small a=4$

$\small \small $a^{2}$=16$

Determine the value of $\small $b^{2}$$

$\small $a^{2}$$+b^{2}$$=c^{2}$$

$\small \small $4^{2}$$+b^{2}$$=6^{2}$$

$\small \small $b^{2}$=20$

Determine the center point to identify the values of h and k. Since the y coordinate of the foci are 4, the center point will be on the same line. Hence, $\small k = 4$ .

Since center point is equal distance from both foci, and we know that the distance between the foci is 12, we can conclude that $\small h=1$

Center point: $\small \small (h, k)= (1,4)$

Thus, the equation of the hyperbola is:

$\small \small $$\frac{(x-1)^{2}$$}{16} - $$\frac{(y-4)^{2}$$}{20}=1$

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Question

Using the information below, determine the equation of the hyperbola.

Foci: and

Eccentricity:

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Answer

General Information for Hyperbola:

Equation for horizontal transverse hyperbola:

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

Distance between foci =

Distance between vertices =

Eccentricity =

$\small $a^{2}$$+b^{2}$$=c^{2}$$

Center: (h, k)

First determine the value of c. Since we know the distance between the two foci is 8, we can set that equal to .

$\small 8=2c$

$\small \small c= 4$

Next, use the eccentricity equation and the value of the eccentricity provided in the question to determine the value of a.

Eccentricity = $\small \small \small \small c/a=2$

$\small \small $\frac{c}{a}$=\frac{2}{1}$=\frac{4}{a}$$

$\small \small 4=2a$

$\small \small a=2$

$\small \small \small $a^{2}$=4$

Determine the value of $\small $b^{2}$$

$\small $a^{2}$$+b^{2}$$=c^{2}$$

$\small \small \small $2^{2}$$+b^{2}$$=4^{2}$$

$\small \small \small $b^{2}$=12$

Determine the center point to identify the values of h and k. Since the y coordinate of the foci are 8, the center point will be on the same line. Hence, $\small \small k = 8$ .

Since center point is equal distance from both foci, and we know that the distance between the foci is 8, we can conclude that $\small \small h=5$

Center point: $\small \small \small (h, k)= (5, 8 )$

Thus, the equation of the hyperbola is:

$\small \small \small \small \small \small $$\frac{(x-5)^{2}$$}{4} - $$\frac{(y-8)^{2}$$}{12}=1$

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Question

Find the center and the vertices of the following hyperbola:

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Answer

In order to find the center and the vertices of the hyperbola given in the problem, we must examine the standard form of a hyperbola:

The point (h,k) gives the center of the hyperbola. We can see that the equation in this problem resembles the second option for standard form above, so right away we can see the center is at:

In the first option, where the x term is in front of the y term, the hyperbola opens left and right. In the second option, where the y term is in front of the x term, the hyperbola opens up and down. In either case, the distance tells how far above and below or to the left and right of the center the vertices of the hyperbola are. Our equation is in the first form, where the x term is first, so the hyperbola opens left and right, which means the vertices are a distance to the left and right of the center. We can now calculate by identifying it in our equation, and then go 3 units to the left and right of our center to find the following vertices:

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Question

Find the equations of the asymptotes for the hyperbola with the following equation:

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Answer

For a hyperbola with its foci on the -axis, like the one given in the equation, recall the standard form of the equation:

, where is the center of the hyperbola.

Start by putting the given equation into the standard form of the equation of a hyperbola.

Group the terms together and terms together.

Factor out from the terms and from the terms.

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

Add to both sides of the equation:

Divide both sides by .

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

The slopes of this hyperbola are given by the following:

$m=\pm$\frac{b}{a}$$

For the hyperbola in question, and .

Thus, the slopes for its asymptotes are $\pm$\frac{4}{6}$=\pm$\frac{2}{3}$$ .

Now, plug in the center of the hyperbola, into the point-slope form of the equation of a line to get the equations of the asymptotes.

For the first equation,

$y-2=\frac{2}{3}$(x+1)$

$y-2=\frac{2}{3}$x+\frac{2}{3}$$

$y=\frac{2}{3}$x+\frac{8}{3}$$

For the second equation,

$y-2=-$\frac{2}{3}$(x+1)$

$y-2=-$\frac{2}{3}$x-$\frac{2}{3}$$

$y=-$\frac{2}{3}$x+\frac{4}{3}$$

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Question

Find the foci of the hyperbola with the following equation:

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Answer

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

and

, where the centers of both hyperbolas are .

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

Group the terms together and the terms together.

Next, factor out from the terms and from the terms.

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

Subtract from both sides.

Divide both sides by .

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

When the term with is first, that means the foci will lie on a horizontal transverse axis.

When the term with is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

For a hyperbola with a horizontal transverse access, the foci will be located at and .

For a hyperbola with a vertical transverse access, the foci will be located at and .

For the given hypebola in the question, the transverse axis is horizontal and its center is located at .

Next, find .

$c=$\sqrt{12+4}$=$\sqrt{16}$=4$

The foci are then located at and .

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Question

Find the foci of the hyperbola with the following equation:

Tap to reveal answer

Answer

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

and

, where the centers of both hyperbolas are .

When the term with is first, that means the foci will lie on a horizontal transverse axis.

When the term with is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

For a hyperbola with a horizontal transverse access, the foci will be located at and .

For a hyperbola with a vertical transverse access, the foci will be located at and .

For the given hypebola in the question, the transverse axis is horizontal and its center is located at .

Next, find .

$c=$\sqrt{27+9}$=$\sqrt{36}$=6$

The foci are then located at and .

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Question

Find the foci of a hyperbola with the following equation:

Tap to reveal answer

Answer

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

and

, where the centers of both hyperbolas are .

When the term with is first, that means the foci will lie on a horizontal transverse axis.

When the term with is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

For a hyperbola with a horizontal transverse access, the foci will be located at and .

For a hyperbola with a vertical transverse access, the foci will be located at and .

For the given hypebola in the question, the transverse axis is horizontal and its center is located at .

Next, find .

$c=$\sqrt{64+36}$=$\sqrt{100}$=10$

The foci are then located at and .

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Question

Find the foci of the hyperbola with the following equation:

Tap to reveal answer

Answer

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

and

, where the centers of both hyperbolas are .

When the term with is first, that means the foci will lie on a horizontal transverse axis.

When the term with is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

For a hyperbola with a horizontal transverse access, the foci will be located at and .

For a hyperbola with a vertical transverse access, the foci will be located at and .

For the given hypebola in the question, the transverse axis is horizontal and its center is located at .

Next, find .

$c=$\sqrt{16+9}$=$\sqrt{25}$=5$

The foci are then located at and .

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Question

Find the foci of the hyperbola with the following equation:

Tap to reveal answer

Answer

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

and

, where the centers of both hyperbolas are .

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

Group the terms together and the terms together.

Next, factor out from the terms and from the terms.

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

Subtract from both sides.

Divide both sides by .

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

When the term with is first, that means the foci will lie on a horizontal transverse axis.

When the term with is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

For a hyperbola with a horizontal transverse access, the foci will be located at and .

For a hyperbola with a vertical transverse access, the foci will be located at and .

For the given hypebola in the question, the transverse axis is horizontal and its center is located at .

Next, find .

$c=$\sqrt{14+50}$=$\sqrt{64}$=8$

The foci are then located at and .

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Question

Which of the following correctly describes the hyperbola of the equation

$$\frac{(x- $2)^{2}$$}{64} - $$\frac{(y-6)^{2}$$}{49} = 1$ .

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Answer

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

is the standard form of a horizontal hyperbola with center . Set ; this hyperbola has its center at .

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Question

Using the information below, determine the equation of the hyperbola.

Foci: and

Eccentricity: $\frac{3}{2}$

Tap to reveal answer

Answer

General Information for Hyperbola:

Equation for horizontal transverse hyperbola:

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

Distance between foci =

Distance between vertices =

Eccentricity =

$\small $a^{2}$$+b^{2}$$=c^{2}$$

Center: (h, k)

First determine the value of c. Since we know the distance between the two foci is 12, we can set that equal to .

$\small c= 6$

Next, use the eccentricity equation and the value of the eccentricity provided in the question to determine the value of a.

Eccentricity = $\small \small c/a=3/2$

$\small $\frac{c}{a}$=\frac{3}{2}$=\frac{6}{a}$$

$\small 12=3a$

$\small a=4$

$\small \small $a^{2}$=16$

Determine the value of $\small $b^{2}$$

$\small $a^{2}$$+b^{2}$$=c^{2}$$

$\small \small $4^{2}$$+b^{2}$$=6^{2}$$

$\small \small $b^{2}$=20$

Determine the center point to identify the values of h and k. Since the y coordinate of the foci are 4, the center point will be on the same line. Hence, $\small k = 4$ .

Since center point is equal distance from both foci, and we know that the distance between the foci is 12, we can conclude that $\small h=1$

Center point: $\small \small (h, k)= (1,4)$

Thus, the equation of the hyperbola is:

$\small \small $$\frac{(x-1)^{2}$$}{16} - $$\frac{(y-4)^{2}$$}{20}=1$

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Question

Using the information below, determine the equation of the hyperbola.

Foci: and

Eccentricity:

Tap to reveal answer

Answer

General Information for Hyperbola:

Equation for horizontal transverse hyperbola:

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

Distance between foci =

Distance between vertices =

Eccentricity =

$\small $a^{2}$$+b^{2}$$=c^{2}$$

Center: (h, k)

First determine the value of c. Since we know the distance between the two foci is 8, we can set that equal to .

$\small 8=2c$

$\small \small c= 4$

Next, use the eccentricity equation and the value of the eccentricity provided in the question to determine the value of a.

Eccentricity = $\small \small \small \small c/a=2$

$\small \small $\frac{c}{a}$=\frac{2}{1}$=\frac{4}{a}$$

$\small \small 4=2a$

$\small \small a=2$

$\small \small \small $a^{2}$=4$

Determine the value of $\small $b^{2}$$

$\small $a^{2}$$+b^{2}$$=c^{2}$$

$\small \small \small $2^{2}$$+b^{2}$$=4^{2}$$

$\small \small \small $b^{2}$=12$

Determine the center point to identify the values of h and k. Since the y coordinate of the foci are 8, the center point will be on the same line. Hence, $\small \small k = 8$ .

Since center point is equal distance from both foci, and we know that the distance between the foci is 8, we can conclude that $\small \small h=5$

Center point: $\small \small \small (h, k)= (5, 8 )$

Thus, the equation of the hyperbola is:

$\small \small \small \small \small \small $$\frac{(x-5)^{2}$$}{4} - $$\frac{(y-8)^{2}$$}{12}=1$

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Question

Find the center and the vertices of the following hyperbola:

Tap to reveal answer

Answer

In order to find the center and the vertices of the hyperbola given in the problem, we must examine the standard form of a hyperbola:

The point (h,k) gives the center of the hyperbola. We can see that the equation in this problem resembles the second option for standard form above, so right away we can see the center is at:

In the first option, where the x term is in front of the y term, the hyperbola opens left and right. In the second option, where the y term is in front of the x term, the hyperbola opens up and down. In either case, the distance tells how far above and below or to the left and right of the center the vertices of the hyperbola are. Our equation is in the first form, where the x term is first, so the hyperbola opens left and right, which means the vertices are a distance to the left and right of the center. We can now calculate by identifying it in our equation, and then go 3 units to the left and right of our center to find the following vertices:

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Question

Find the equations of the asymptotes for the hyperbola with the following equation:

Tap to reveal answer

Answer

For a hyperbola with its foci on the -axis, like the one given in the equation, recall the standard form of the equation:

, where is the center of the hyperbola.

Start by putting the given equation into the standard form of the equation of a hyperbola.

Group the terms together and terms together.

Factor out from the terms and from the terms.

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

Add to both sides of the equation:

Divide both sides by .

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

The slopes of this hyperbola are given by the following:

$m=\pm$\frac{b}{a}$$

For the hyperbola in question, and .

Thus, the slopes for its asymptotes are $\pm$\frac{4}{6}$=\pm$\frac{2}{3}$$ .

Now, plug in the center of the hyperbola, into the point-slope form of the equation of a line to get the equations of the asymptotes.

For the first equation,

$y-2=\frac{2}{3}$(x+1)$

$y-2=\frac{2}{3}$x+\frac{2}{3}$$

$y=\frac{2}{3}$x+\frac{8}{3}$$

For the second equation,

$y-2=-$\frac{2}{3}$(x+1)$

$y-2=-$\frac{2}{3}$x-$\frac{2}{3}$$

$y=-$\frac{2}{3}$x+\frac{4}{3}$$

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Question

Find the foci of the hyperbola with the following equation:

Tap to reveal answer

Answer

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

and

, where the centers of both hyperbolas are .

When the term with is first, that means the foci will lie on a horizontal transverse axis.

When the term with is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

For a hyperbola with a horizontal transverse access, the foci will be located at and .

For a hyperbola with a vertical transverse access, the foci will be located at and .

For the given hypebola in the question, the transverse axis is horizontal and its center is located at .

Next, find .

$c=$\sqrt{27+9}$=$\sqrt{36}$=6$

The foci are then located at and .

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Question

Find the foci of a hyperbola with the following equation:

Tap to reveal answer

Answer

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

and

, where the centers of both hyperbolas are .

When the term with is first, that means the foci will lie on a horizontal transverse axis.

When the term with is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

For a hyperbola with a horizontal transverse access, the foci will be located at and .

For a hyperbola with a vertical transverse access, the foci will be located at and .

For the given hypebola in the question, the transverse axis is horizontal and its center is located at .

Next, find .

$c=$\sqrt{64+36}$=$\sqrt{100}$=10$

The foci are then located at and .

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Question

Find the foci of the hyperbola with the following equation:

Tap to reveal answer

Answer

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

and

, where the centers of both hyperbolas are .

When the term with is first, that means the foci will lie on a horizontal transverse axis.

When the term with is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

For a hyperbola with a horizontal transverse access, the foci will be located at and .

For a hyperbola with a vertical transverse access, the foci will be located at and .

For the given hypebola in the question, the transverse axis is horizontal and its center is located at .

Next, find .

$c=$\sqrt{16+9}$=$\sqrt{25}$=5$

The foci are then located at and .

← Didn't Know|Knew It →

Question

Find the foci of the hyperbola with the following equation:

Tap to reveal answer

Answer

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

and

, where the centers of both hyperbolas are .

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

Group the terms together and the terms together.

Next, factor out from the terms and from the terms.

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

Subtract from both sides.

Divide both sides by .

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

When the term with is first, that means the foci will lie on a horizontal transverse axis.

When the term with is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

For a hyperbola with a horizontal transverse access, the foci will be located at and .

For a hyperbola with a vertical transverse access, the foci will be located at and .

For the given hypebola in the question, the transverse axis is horizontal and its center is located at .

Next, find .

$c=$\sqrt{14+50}$=$\sqrt{64}$=8$

The foci are then located at and .

← Didn't Know|Knew It →

Question

Find the foci of the hyperbola with the following equation:

Tap to reveal answer

Answer

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

and

, where the centers of both hyperbolas are .

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

Group the terms together and the terms together.

Next, factor out from the terms and from the terms.

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

Subtract from both sides.

Divide both sides by .

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

When the term with is first, that means the foci will lie on a horizontal transverse axis.

When the term with is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

For a hyperbola with a horizontal transverse access, the foci will be located at and .

For a hyperbola with a vertical transverse access, the foci will be located at and .

For the given hypebola in the question, the transverse axis is horizontal and its center is located at .

Next, find .

$c=$\sqrt{12+4}$=$\sqrt{16}$=4$

The foci are then located at and .

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Question

Which of the following correctly describes the hyperbola of the equation

$$\frac{(x- $2)^{2}$$}{64} - $$\frac{(y-6)^{2}$$}{49} = 1$ .

Tap to reveal answer

Answer

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

is the standard form of a horizontal hyperbola with center . Set ; this hyperbola has its center at .

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