1
$\begingroup$

Given two hyperbolas h1(with foci and center) and h2(with foci and center), In what condition these hyperbolas will not intersect to each other?. I can get the condition when h1 and h2 are standard hyperbola (parallel to axis and the center is the origin). I want to find the condition when both of them are not standard hyperbola. Thanks

  • 0
    I think you ask a too general question. You should for example restrict your attention to equilateral hyperbolas and take one of them as the standard equilateral hyperbola with equation $xy=1$ and the other one as a rotated + translated image of the first one. Besides, do you know the focal definition $|MF-MF'| \ = \ $constant of a hyperbola ? – 2017-02-16
  • 0
    @JeanMarie. Thanks and apologize for unclear question. This question is for equilateral hyperbola. I know that constant property of hyperbola – 2017-02-16

3 Answers 3

2

Let $x^2-y^2=1$ be the first hyperbola.

  • Re-scaling $$x^2-y^2=c^2 \tag{$c^2 \ne 1$}$$

  • Conjugate $$y^2-x^2=c^2$$

  • Translation \begin{align*} (x-h)^2-(y-k)^2 &= 1 \\ 2(ky-hx)+h^2-k^2 &= 0 \\ x &= \frac{h^2-k^2+2ky}{2h} \\ (h^2-k^2+2ky)^2-4h^2 y^2 &= 4h^2 \\ 4(h^2-k^2)y^2-4k(h^2-k^2)y+4h^2-(h^2-k^2)^2 &= 0 \\ \Delta & < 0 \\ k^2(h^2-k^2)^2-(h^2-k^2)[4h^2-(h^2-k^2)^2] & < 0 \\ h^2(h^2-k^2)(h^2-k^2-4) &< 0 \\ \end{align*}

$$\fbox{$0

enter image description here

  • 0
    Thanks. I still don't get the second line (before x=......). Where are x^2 and y^2 go, they suddenly disappear. – 2017-02-17
  • 0
    Equating $x^2-y^2=(x-h)^2-(y-k)^2=1$, all quadratic terms have gone. Then substitute either back into either hyperbola. – 2017-02-17
  • 0
    [+1] Good treatment of the condition $0 – 2017-02-18
  • 0
    In general, if conic $ax^2+bxy+cy^2=1$ is translated by $(h,k)$ such that they don't intersect, then $$(b^2-4ac)(ah^2+bhk+ck^2)(ah^2+bhk+ck^2-4)<0$$ – 2017-02-18
  • 0
    In general hyperbola equation ax^2+bxy+cy^2=1. Are a, b and c here similar with a,b,c in standard hyperbola equation x^2/a^2 - y^2/b^2 = 1, where a^2+b^2 = c^2? Thanks – 2017-02-20
  • 0
    No, it's general form for central conics. Some texts may write it in quadratic form: $$\begin{pmatrix} x & y \end{pmatrix} \begin{pmatrix} a & \frac{b}{2} \\ \frac{b}{2} & c \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}=1$$ – 2017-02-20
0

use the answer of this question to get two equations, set them inequal to each other, then that is your condition, swap them back to focal model.

PS: it might be easier to answer in polar or bipolar coordinates, as they seem more suitable for parabola solving problems

0

Let us consider the reference equilateral hyperbola with equation $xy=1$ (see blue curve on graphics below). Any rectangular hyperbola can be obtained from the reference with a rotation (angle $\theta$) followed by a translation (vector $\binom{a}{b}$). One should obtain the following result :

  • if $\theta\neq0$, there are intersection points.

  • If $\theta=0$, the only translations that give no intersection point are with vector $k\binom{1}{1}$ for $-2

I thought I have a proof but it is not the case.

enter image description here

  • 0
    I have withdrawn the major part of my "proof" (which had a flaw). and modied the graphics. – 2017-02-16