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## Precalculus (2018 edition)

### Course: Precalculus (2018 edition)>Unit 2

Lesson 10: Hyperbolas not centered at the origin

# Equation of a hyperbola not centered at the origin

Sal analyzes the hyperbola whose equation is (x-1)^2/16-(y+1)^2/4=1, and graphs it. Created by Sal Khan.

## Want to join the conversation?

• One thing that is a bit unclear to me in these types of equations: At Sal has (x-1)^2 /16 - (y+1)^2/4 = 1 so the center is (1,-1). But if you plug both those coordinates back into the equation you get 0 = 1 which makes no sense. What detail am I missing here? •   Bill,

That's an excellent observation, and you'll find the same thing happens in the equations of the circle and the ellipse, too. So let's use the equation of a circle first, and I think we'll talk about what happens with it… and I think that will help make the hyperbola make more sense too.

So here is an equation of a circle:
(x-h)^2+(y-k)^2=r^2

In this equation the center of the circle is at (h,k), and the circle has a radius equal to r. So let's throw in some numbers really quick:

(x-3)^2+(y-4)^2=25

Ok, in this circle, the center is at (3,4) and the radius is 5, because r^2 is 25 and 5 is the square root of 25. Now, notice what happens if you plug in 3 for x and 4 for y? You get:
(3-3)^2 + (4-4)^2 =25
or: (0)^2 +(0)^2 =25
or: 0=25

Now, clearly 0 doesn't equal 25, and it's the same problem that you noticed on the hyperbola. So what is going on?

Well, remember that the point (3,4) is the CENTER of the circle, and the equation gives us the formula for the circle itself. The center of the circle isn't actually ON the circle, though, which is why when we plug in the coordinates for the center of the circle the equation doesn't evaluate to a true statement. The equation ONLY evaluates to a true statement if we plug in points that are ON the circle. If we plug in any points that aren't on the circle, the equation doesn't evaluate to a true statement.

The same thing happens in the equation for a hyperbola. The center of the hyperbola isn't actually on either of the two curves that make up the hyperbola. So if we plug in the numbers for the center of the hyperbola, the equation doesn't evaluate to a true statement. That doesn't mean that we don't have the correct numbers for the center… it just means that the center of the hyperbola isn't actually on the curves that make up the hyperbola just like the center of a circle isn't on the curve that draws the circle.

Does that help this make sense?
• How do you know where the foci are on the graph? • Can somebody say what the equation to find the asymptotes is? He never really says, he does this in a really weird manner that for me at least is really hard to understand, i'd rather it be shown more simply • Why does that make a difference when x approaches infinity we forget about the +or- 4 for us now that x is basically in the area of + or - 10 for arguments sake +or - 4 should make a HUGE difference? • When X = 10, yes, the -4 under the square root sign does make a big difference. However, for argument's sake, let's say that X = 1,000 in the equation from the video y = +or- sqrt(x^2 /4 -4) Then: sqrt (1,000,000/4 - 4) which can simplify to +or- sqrt (249,996) or +or- 499.996 If you take away the -4 from under the square root, you get +or- sqrt (250,000) or +or- 500 which gives a difference of .004. A very small difference, and that's just at X = 1,000. Imagine how small the difference is at X = 1,000,000 or X = 100,000,000,000,000...
• Is it possible for a hyperbola to have no x and y intercepts whatsoever? • isnt the asymptote of the real equation slightly different than the one he wrote? • I'm a bit confused, What exactly is he solving for when he gets to ? And what exactly did he do by getting there? • I am trying to understand why the hyperbolas formula is equal to 1? Why not for example equal to 0!
(1 vote) • Well, the standard formula for the hyperbola is an equation, so if it is a number not equal to 0 then you can just divide by that number on both sides to simplify the equation to the point where it does equal 1.
And when the formula is equal to 0, you actually get the asymptotes of the hyperbola! The hyperbola equation equal to 0 can be shown as
(x^2)/(a^2)-(y^2)/(b^2)=0
where any transformations are applied to the asymptotes as well. Let's try simplifying it!
Add to both sides and get:
(x^2)/(a^2)=(y^2)/(b^2)
Take the square root:
x/a= plus or minus y/b
multiply by b (or -b) on both sides:
y= plus or minus bx/a
which can be written as
f(x)=plus or minus (b/a)x
which is actually the equation(s) for the asymptotes!  