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View Full Version : How do I figure out what conic section this equation makes?


maramfaith24
Jan 9, 2011, 03:26 PM
I'm not looking for answer, just the step to finish the equation and figure out which conic section it forms?

Here's the equation: x^2=y^2+9.

I think I start by putting the equation into y=mx+b form, so it can be graph on a graphing calculator.

But I can't figure out how it's going to look in the end, can you help?

galactus
Jan 9, 2011, 03:44 PM
How could you write it in y=mx+b form? y=mx+b is the slope intercept form of a line.

This has nothing to do with a line.

I think you are just referring to solving it for y so you can graph it.

Yes, you can do that. y=\pm \sqrt{x^{2}-9}

To solve it without graphing, rewrite it as:

x^{2}-y^{2}=9

Divide by 9:

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

This is a hyperbola with vertices at (-3,0) and (3,0)

maramfaith24
Jan 9, 2011, 04:35 PM
So would re-write be the easier way of the two? I'm sorry, I am just really conused. I'm an online student, so if I don't understand something.. This website is normally where I come.

maramfaith24
Jan 9, 2011, 04:49 PM
The way my teacher wants my to do it, is to first solve or y.

galactus
Jan 10, 2011, 05:42 AM
I already solved it for y.

Graph that in your calculator.

Don't forget the positive and negative or you'll just get one side.

Graph y=\sqrt{x^{2}-9} and y=-\sqrt{x^{2}-9}