Hyperbolas Examples

Find the center, vertices, foci, and asymptotes of .

We have a hyperbola in the form .

It's center is at (h, k), or (-6, 6). The values a and b, which will be very helpful for finding everything else, are 4 and 6, respectively. Remember, a always hangs with x, and b is always under y. Which one is positive or negative doesn't change that a bit.

Hyperbolas have two vertices, and their locations depend on if it is a vertical or horizontal hyperbola. And that depends on which variable is positive. Here, y is positive, so the graph will be vertical. The vertices will be above and below the center by b. That is, (-6, 6 ± 6) = (-6, 0) and (-6, 12).

Let's nab the two asymptotes next. They are identical, except that the signs of their slopes are opposite. Their equations are .

Be careful with h and k. We put -h next to x, but positive k outside the parentheses.

Last but probably not least are the foci. We've got an equation to find their distance from the center, and it is f² = a² + b². Plugging in our values, we get:

f² = 16 + 36 = 52

That's just the distance to the foci; we need to find their coordinates to finish the problem. They go in the same directions from the center as the vertices do, so they are above and below the as well.

(-6, -1.21) and (-6, 13.21)

We've finally found all the important parts of this hyperbola.

Find the center, vertices, foci, and asymptotes of .

This is a horizontal hyperbola—it opens to the left and right. The transverse axis is a horizontal line, and the center, vertices, and foci will only differ in their x-coordinate. That's quite a lot of information just from looking at the first term in the equation.

The center? That's (1, 2). a and b? They're 6 and 9. The meaning of life? We hear that "42" is a popular answer, but we think "ice cream" is a strong contender.

The vertices will only change in x, so that's a's domain. (1 ± 6, 2) = (-5, 2) and (7, 2).

Calculating the foci distance is next. Plugging into f² = a² + b², we get:

f² = 36 + 81 = 117

Going back to the center, we can move left and right by 10.82 to get the coordinates for the foci:

(-9.82, 2) and (11.82, 2)

All that's left are the pens that keep our hyperbola in check, the asymptotes. Plugging a, b, h, and k into their formula gets us:

Center, vertices, foci, asymptotes? Check, check, check, check.

Find the equation of the horizontal hyperbola that has:

Asymptote: 

This is pretty slim picking for figuring out the whole hyperbola's equation. We only know 1) that the hyperbola is horizontal, so x is the positive term, and B) one of the two asymptotes.

That's enough, though, because that asymptote gives us the center, as well as a and b. Once we have those, we are in business.

For the center, h is 2, like we expect. Be careful with k, though. That positive 4 is actually a positive 4, just as it appears; we don't switch the sign like we have been recently. Remember, this is just a plain-Jane line we're dealing with, so it passes through (2, 4), not (2, -4).

a and b come right from the slope of the asymptote. b, the term associated with y, is on top, and a, x's buddy, is on bottom. It all makes sense. b = 3, a = 2. Now we bring it all together into the equation of our dreams.