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How to prove $\frac{y^2-x^2}{x+y+1}=\pm1$ is a hyperbola, knowing the canonical form is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=\pm1$ where $a$ and $b$ are constants? Thanks !

Shuchang
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user111288
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2 Answers2

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Taking the '+' sign, $$y^2-x^2=x+y+1\implies\left(y-\frac12\right)^2-\left(x+\frac12\right)^2=1^2$$

$$\implies\frac{\left(y-\frac12\right)^2}{1^2}-\frac{\left(x+\frac12\right)^2}{1^2}=1^2$$

Similarly for the '-' sign,

$$y^2-x^2=-(x+y+1)\implies\left(x-\frac12\right)^2-\left(y+\frac12\right)^2=1^2$$

Can you recognize $a,b$ here?

lab bhattacharjee
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  • I trully appreciate your reply, but hardly see how this helps since, no, I can't recognize $a$ and $b$... – user111288 Nov 29 '13 at 04:47
  • @user111288, how about the edited version? See also, http://mathworld.wolfram.com/RectangularHyperbola.html – lab bhattacharjee Nov 29 '13 at 04:49
  • Trying as hard as I can to, but no...! One more edit, maybe? :P! Edit : Ha, okay, wait a second, looking at that Wolfram link ! – user111288 Nov 29 '13 at 04:53
  • @user111288 If the center of the hyperbola is at $(h,k)$ then the "canonical form" is $(y-k)^2/a^2 -(x-h)^2/b^2=\pm 1$. [Usually I've seen the $a$ associated to the $x$ term and the $b$ to the $y$ term, however.] To put this another way, the "canonical form" may be for a different coordinate system than that of the originally given conic, which means if originally one has an equation in $x,y$ one might need new variables $u,v$ for the "standard form". This is especially true if a rotation is needed. – coffeemath Nov 29 '13 at 04:56
  • Hah ! THANK YOU very much ! The hyperbola isn't centered at (0,0) ! – user111288 Nov 29 '13 at 04:57
  • @user111288, As the transformation of axes does not change the nature of a curve, we can set $y-\frac12=Y,x+\frac12=X$ in the first case – lab bhattacharjee Nov 29 '13 at 04:59
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Let $$ \frac{y^2-x^2}{x+y+1}=1\\ \Rightarrow y^2-x^2=x+y+1\\ \Rightarrow y^2-x^2-x-y=1 $$ Complete the squares for x and y . You will get rectangular hyperbola. Similar will be the case if $$ \frac{y^2-x^2}{x+y+1}=-1$$

GTX OC
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