Results 1 to 5 of 5

- Oct 22nd 2011, 10:49 PM #1

- Joined
- Feb 2010
- Posts
- 460
- Thanks
- 34

- Oct 23rd 2011, 05:57 AM #2

- Oct 23rd 2011, 07:41 AM #3

- Joined
- Feb 2010
- Posts
- 460
- Thanks
- 34

- Oct 23rd 2011, 09:07 AM #4

- Joined
- Apr 2005
- Posts
- 19,729
- Thanks
- 3010

## Re: area4

If you add and subtract $\displaystyle y^2$ you get $\displaystyle x^2+ 2xy+ y^2- y^2+ 40|x|= (x+y)^2- y^2+ |x|= 400$. Let u= x+y and v= y. x= u- y= u- v so the equation becomes $\displaystyle u^2- v^2+ 40|u- v|= 400$.

For x= u- v>0, |x|= |u-v|= u- v so that becomes $\displaystyle u^2+40u- v^2- 40v= 400$. Complete the square by adding 400 and subtracting 400 on the left: $\displaystyle u^2+ 40u+ 400- (v^2+ 40v- 400)= (u+20)^2- (v+20)^2= 400$. That is a hyperbola, in the uv-plane, with center at (-20, -20), vertices (0, -20) and (-40, -20), and vertical asymptotes at 45 degree angles to the axes. That is, one of the asymptotes**is**the line u= v and, since we require that u- v> 0 here, we have only the branch of the hyperbola on that side.

If u- v< 0, then |u- v|= v- u> 0 and the equation becomes $\displaystyle u^2- 40u- v^2+ 40v= 400$ and, after completing the square, $\displaystyle (u- 20)^2- (v- 20)^2= 400$, a hyperbola with center at (20, 20), vertices (0, 20) and (40, 20), and u= v and u= -v as asymptotes again. Since u< v now, we have the branch of the hyperbola now on the opposite side. Those don't cross: one lies on one side of v= u and the other on the opposite side, so I don't see how they can create a bounded region!

- Oct 24th 2011, 12:38 AM #5