# Homework Help: Hyperbola's asymptote, where's my mistake?

1. Jul 6, 2016

### Poznerrr

• Member warned to not delete the homework template in future posts
I have tried like 5 times to do this problem ans still don't get the answer I'm supposed to (A). Anybody finds mistake in my work below?

2. Jul 6, 2016

### Fightfish

If I'm not mistaken, it should be $a/b$ and not $b/a$. so the factor appearing will be $2/3$ and not $3/2$.

3. Jul 6, 2016

### Poznerrr

Yeah, I thought about it, and it works out with a/b being the slope of the asymptote. But different sources stick to either a/b and b/a, so it's really confusing which slope is actually correct.

4. Jul 6, 2016

### Fightfish

The way to remember it is well, to derive it yourself, which isn't very hard. The general equation for a hyperbola is in the form
$$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$$
In the asymptotic limit, the term $1$ becomes insignificant and so we drop it, leaving
$$\frac{y^2}{a^2} = \frac{x^2}{b^2}$$
and so
$$y = \pm \frac{a}{b} x$$

5. Jul 6, 2016

### micromass

Given a conic section
$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$
the asymptotes can be calculated by a little trick. Basically, the slope of the asymptote is $s$, where $s$ is the solution to $A + Bs + Cs^2=0$.

In your case, this corresponds to the solution of $4-9s^2=0$. So we get $s = 2/3$ or $s= -2/3$. The slope of (A) is $-2/3$.

6. Jul 6, 2016

### Math_QED

Ah, the trick with the homogeneous coordinates and equations ;)

7. Jul 6, 2016

### Poznerrr

Thanks! I'll use this trick in future.

8. Jul 7, 2016

### micromass

You shouldn't unless you can prove the trick actually works.

9. Jul 7, 2016

### Irene Kaminkowa

Poznerrr,
For the proof, divide the initial conic section by x2. And find the limit if x->infinity. Thus, your get the equation from post #5, where s = y/x, that is the slope.

10. Jul 7, 2016

### Ray Vickson

Also: in the future do not post images of your work, as most people on this Forum will not read them and will not be willing to help you. Just type out your work.