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The hyperbola being an orthogonal parabola, for which $(-1,2)$ is a focal point and $x-y+1=0$ is an asymptote.

If I have the equation for the asymptote $y=x+1$ is the center $(0,1)$?

I do not know where to proceed next.

mathematics2x2life
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Jake
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1 Answers1

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No. The center is $(-1,0)$.

Since the hyperbola being an orthogonal parabola, we have $$\frac{(x-p)^2}{a^2}-\frac{(y-q)^2}{b^2}=-1.$$ Since $(−1,2)$ is a focal point, we have $$-1=p, 2=\sqrt{a^2+b^2}+q.$$ Since $x−y+1=0\iff y=x+1$ is an asymptote, we have $$1=\frac ba, 1=-\frac{bp}{a}+q$$ Now we have $$q=0, p=-1, a=b, a^2=2.$$ Hence, the center is $(p,q)=(-1,0)$.

mathlove
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  • I think it's good to leave some spaces for OP to give his/her own try. – Shuchang Dec 29 '13 at 01:55
  • You are right. I'll do so. – mathlove Dec 29 '13 at 01:58
  • I do not understand how you got -1=p, 2=sqr(a^2+b^2)+q? Is there a special formula for this? – Jake Dec 29 '13 at 11:42
  • and also the part 1=-bp/a + q ? – Jake Dec 29 '13 at 11:43
  • If $(x^2)/(a^2)+(y^2)/(b^2)=-1,$ then focal points are $(0,\pm\sqrt{a^2+b^2})$ and its asymptotes are $y=\pm\frac ba x.$ Now we translates its center from $(0,0)$ to $(p,q)$, the focal points are $(p, \sqrt{a^2+b^2}+q)$, and its asymptotes are $y-q=\pm\frac ba (x-p).$ – mathlove Dec 29 '13 at 12:06
  • Okay so I do get it now. I get b^2=2 as a and b are equal. So the equation of the hyperbola is y^2/2 - (x+1)^2/2=1. However there are two solutions. The other one is (x-1)^2/2 - (y-2)^2/2=1 but I do not know how they get it. Does it have to do something with +/- b/a? – Jake Dec 29 '13 at 13:20
  • I understand what you think. At first, I set $(x-p)^2/a^2+(y-q)^/b^2=1$, but I faced some strange things. Sorry, but I forgot the details. Just try the case, you'll face something strange. – mathlove Dec 29 '13 at 14:43