# Find the length of the Transverse axis. What did I do wrong?

1. Mar 8, 2005

### trigger352

I'm doing an Exam correction an I can't see how this answer is found.

Question: $$x^2 - \frac{\ y^2}{9} = 1$$
Find the length of the transverse axis.

I took $$x^2 - \frac{\ y^2}{9} = 1$$

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

to get

$$\frac{\ (x-0)^2}{1^2} - \frac{\ (y-0)^2}{3^2} = 1$$

The 0's were because in the original equation, I didn't see an $$h$$ or $$k$$ value that was affecting either $$x$$ or $$y$$
The $$1^2$$ is because $$x$$ = $$\frac{x}{1}$$ like $$a$$ = $$\frac{a}{1}$$

$$a = 3$$, therefore the length of the transverse is 6 units...

2. Mar 8, 2005

### HallsofIvy

Staff Emeritus
When x= 0 what is y?

Setting x= 0 gives $-\frac{y^2}{9}= 1$- which has NO solution.

When y= 0 what is x?

Setting y= 0 gives x2= 1 so x= -1 and 1. THAT'S the "transverse axis" you want!