Finding X and Y intercepts - Hyperbola

K.ourt

New member
Joined
Jul 20, 2005
Messages
17
Ok, Im supposed to find the x and y intercepts of this hyperbola that I have drawn, but im running into some minor issues here. I know you're supposed to make y and x equal to zero (not at the same time obviously) and Ive tried that, but wouldnt that only give me one intercept? Oye! Here's the question:
General Form: 16x^2 - y^2 - 96x + 8y +112 = 0
Standard: ((x-3)^2)/1 - ((y-4)^2)/16 = 1

so the center is (3,4) therefore its not centered at (0,0), which is what I think is giving me trouble, since all the websites Ive looked at for help, and in my notes, has the hyperbola centered at (0,0). It opens left/right. Vertices at (2,4) and (4,4). And it crosses the x-axis twice as well as the y-axis twice. I just need help, or a hint, or even a website to look at, that will help. Im so confused. Heh. Oh, and in the text book it says things like Focus and Directrix, but our teacher told us not to look in the text for help, because we aren't dealing with those words. So yea..
 
All you're being asked for is the intercepts. You don't need to find the center, the foci, the vertices, the axis lengths, or any of that. Just follow the usual intercept procedure:

. . . . .y-intercept(s): plug "0" in for "x", and solve

. . . . .x-intercept(s): plug "0" in for "y", and solve

Eliz.
 
Oh, so it still does work that way. Im obviously having a moment. Oye. Sorry for all the babble, and wasted time. Ive gotten the x intercepts, now Im working on the y.
 
K.ourt said:
Oh, so it still does work that way.
Always.

Don't let the particular context trip you up. In the x,y-plane, you always find the intercepts on the one axis by plugging in zero for the other variable. :D

Eliz.
 
Top