# Hyperbola problem

1. Dec 19, 2009

### farmd684

1. The problem statement, all variables and given/known data
Find an equation of a hyperbola that passes through origin and has asymptotes y = 2x+1 and y= -2x+3

2. Relevant equations

3. The attempt at a solution

I have got the center ( 1/2,2 ) and as it passes through the center i have this equation

4/a2 - 1/(4b2) from this information how can i get the ratio a/b ?

Thanks

2. Dec 19, 2009

### Mentallic

Draw a diagram first (with reasonable accuracy). Notice that to have the hyperbola pass through the origin with those asymptotes, it must be an "up-down" hyperbola. This means the standard form for the hyperbola that youve used needs to be equal to -1 rather than 1.

So you have $$\frac{4}{a^2}-\frac{1}{4b^2}=-1$$

And it isn't possible to get the ratio a/b out of this equation. You need to use another piece of information. For a standard hyperbola that has the centre at the origin, what are the equations of its asymptotes?

3. Dec 19, 2009

### farmd684

I know the fact that it is a up down hyperbola. considering that info and the coordinate of the center i have this form of equation :

(y-2)2 /a2 - (x-1/2)2/b2 = 1and putting (0,0) in this equation i get this form

4/a2 - 1/(4b2) = 1

so i guess it should be right. I have to find now the ratio a/b. And the general form of asymptote lines for a hyperbola in standard position is

y-k = +- b/a (x-h) where h,k is the coordinate of the center.

4. Dec 19, 2009

### Mentallic

Ahh yes sorry for that. Long, embarrassing story short, I mixed up the b and a.

Ok so what is the ratio of b/a then? If you're unsure, notice the similarity between that formula, and the point-gradient formula for a line:

$$y-k=\pm \frac{b}{a}(x-h)$$

$$y-y_1=m(x-x_1)$$

Now when you find the ratio of b/a, you can simultaneously solve that with the other equality to find both a and b.

EDIT:
I now realize what the little mix up about how the standard form should be set out was. We were both wrong. Again, I'm sorry, I found conics kind of dodgy when I studied them a while back, so now my memory has made it worse. But trust that I'm right now

The standard side-side hyperbola with centre at (h,k) is $$\left(\frac{x-h}{a}\right)^2-\left(\frac{y-k}{b}\right)^2=1$$ This formula is most important and you can adapt the other types from this one!

But when you're dealing with up-down hyperbolas, it is instead equal to -1.

So what we have when multiplying through by -1 is $$\left(\frac{y-k}{b}\right)^2-\left(\frac{x-h}{a}\right)^2=1$$

See how the b divides the y coordinate, not the x.

Last edited: Dec 19, 2009
5. Dec 20, 2009

### farmd684

so i have to find the gradient of those two lines ? It gives me 2 and -2 for those two lines. Then b/a = +-2 ?

6. Dec 20, 2009

### Mentallic

Yes that's right.

Simultaneously solve

$$\frac{4}{b^2}-\frac{1}{4a^2}=1$$

and

$$\frac{b}{a}=\pm 2$$

But you should notice that that when you square both sides it rids you of the $\pm$ so don't be freaked out by it

7. Dec 20, 2009

### tiny-tim

Hi farmd684! Thanks for the PM.
Start again …

since the centre is (1/2,2), the equation must be (x - 1/2)2/a2 - (y - 2)2/b2 = constant,

so use the ratio a/b from the given asymptotes.

8. Dec 20, 2009

### farmd684

The focal axis should always be defined as (a) in hyperbola (or not). So in my book all up down hyperbola are defined by y2/a2 - x2 /b2 form. I also know that for a updown hyperbola i have asymptotic ratio defined by a/b whereas it is b/a for a right left. As the coordinates are below both of the asymptotic lines so should't it be a updown hyperbola thus following the pattern y2/a2 ?

A similar problem from book which answer is given :

Asymptotes y=x-2 and y=-x+4 and passes through the origin.
center ( 1/2,2)
they have found a/b=2 how they got it ?

9. Dec 20, 2009

### tiny-tim

Hi farmd684!

I don't know why you're following all these rules …

a hyperbola (with asymptotes of equal slope) has to be of the form (x - p)2/a2 - (y - q)2/b2 = constant …

if the constant is positive it'll be up-down, and if it's negative it'll be left-right (or is it the other way round?) …

just put the numbers in and see what works …

the constant has to be such that x = y = 0 (in this case) satisfies the equation, and also such that when x = y = ∞, y = ± 2x.

That's all you need to know!

10. Dec 20, 2009

### HallsofIvy

Look at
$$\frac{x^2}{a^2}- \frac{y^2}{b^}= 1$$
for very large values of x and y. For "regular size" a and b, both $x^2/a^2$ and $y^2/b^2$ will be very, very large. In fact, that poor little "1" on the right is negligible in comparison.

That is, for very large x and y, the graph will be very close to the graph of
$$\frac{x^2}{a^2}- \frac{y^2}{b^2}= \left(\frac{x}{a}- \frac{y}{b}\right)\left(\frac{x}{a}+ \frac{y}{b}\right)= 0$$
which is just the two straight lines corresponding to x/a= y/b and x/a= -y/b or $y= \pm\frac{b}{a}x$.

As to how your book got b/a= 2 from asymptotes y=x-2 and y=-x+4, I have no idea. Since those lines have slope 1 and -1, it should be b/a= 1, not 2. The two lines intersect, of course, where y= x- 2= -x+ 4 so that 2x= 6, x= 3. Then y= 3-2= -3+ 4= 1. If the asymptotes are y= x-2 and y= -x+ 4, as you say, the center is at (3, 1), not (1/2, 2). Such a hyperbola must be of the form $(x-3)^2- (y-1)^2= k$ for some (non-zero) number k. If, in addition to having those asymptotes, the hyperbola passes through the origin, we must have $(-3)^2- (-1)^2= 8= k$ so that k is either 10 or -10 giving either $(x-3)^2- (y-1)^2= 8$ or $(x-3)^2- (y-1)^2= -8$ so that $(y-1)^2- (x-3)^2= 8$.

That is, such a hyperbola must be either
$$\frac{(x-3)^2}{8}- \frac{(y-1)^2}{8}= 1$$
or
$$\frac{(y-1)^2}{8}- \frac{(x-3)^2}{8}= 1$$

11. Dec 20, 2009

### farmd684

Thanks for the explanation
The book definitely published wrong answer.
Well you wrote
so that k is either 10 or -10. It should be 8 or -8 right ?