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}