Hi guys!
use the graph 4x^2-y^2-8x-4y+16=0 to do the following:
a. write the equation in standard form
I did but could you double check my work?
4x^2-y^2-8x-4y+16=0
4x^2-y^2-8x-4y=-16
4(x^2-2x) -(Y^2+4y) =-16
4[(X^2-2x+1)-1] - [(Y^2+4y+4)-4] =-16
4[(x-1)^2-1] - [(Y+2)^2-4] =-16
4(x-1)^2-4 - (y+2)^2+4 =-16
4(x-1)^2- (y+2)^2= -16
4(x-1)^2- (y+2)^2= -16
-16
(x-1)^2 - (y+2)^2 = 1
-4 ....... 16
I think I did that right! but not sure!
there is another part to it:
locate the center and the vertices of this curve
The center is:
(-2,1)
how would I find out the vertices of the curve?
Thanks for your help!