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So the fixed points are $$F_1=(p_1,q_1)$$$$F_2=(p_2, q_2)$$ Mid-point of foci(centre) is $$\left(\cfrac{p_1+p_2}{2},\cfrac{q_1+q_2}{2}\right)=(c_x,c_y)$$ and the the point $P=(h,k)$

The equation is given by: $$|PF_1-PF_2|=d$$ Further solving results in :
Case:1 $$-2(p_1-p_2)h-2(q_1-q_2)k+q_1^2-q_2^2+p_1^2-p_2^2=d^2+2d\sqrt{(p_2-h)^2+(q_2-k)^2}.$$
Case:2
$$-2(p_2-p_1)h-2(q_2-q_1)k+q_2^2-q_1^2+p_2^2-p_1^2=d^2+2d\sqrt{(p_2-h)^2+(q_2-k)^2}.$$

Combining the results
$$(p_1-h)^2+(q_1-k)^2+(p_2-h)^2+(q_2-k)^2-2\left(\sqrt{(p_1-h)^2+(q_1-k)^2}.\sqrt{(p_2-h)^2+(q_2-k)^2}\right)=d^2$$

Is there any further simplification that can be done?

In hindsight this has to be the equation of hyperbola. So I proceed to change it to such.
Let coordinates of the vertices of hyperbola be $(v_x,v_y)$.
Then I think$$\frac{v_y-c_y}{v_x-c_x}=\frac{v_y-q_1}{v_x-p_1}=\frac{q_2-q_1}{p_2-p_1}$$

However I don't know How to proceed further?

This proof assumes the center of hyperbola to be at (0,0) and x-axis to be transverse axis. I was trying for a more general equation.

RobPratt
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Aurelius
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  • While we are at it. Please suggest a link to prove why a hyperbola has a vertex at all ? In other words why is there a point on the line joining the foci the difference of whose distances from the foci is a given constant? – Aurelius Nov 13 '23 at 14:12
  • Also why should the vertex be between the foci. Why can't both the foci be on same side of a vertex? Because without the above assumptions it seems impossible to derive the general/standard equation of hyperbola. But surely they aren't baseless assumptions – Aurelius Nov 13 '23 at 14:18
  • Triangular inequality implies that $|PF_1-PF_2|<F_1F_2$ if $P$ doesn't lie on line $F_1F_2$. On the other hand, $|PF_1-PF_2|=F_1F_2$ if $P$ lies on line $F_1F_2$ outside segment $F_1F_2$. Hence if you want a non-degenerate hyperbola you need $d<F_1F_2$ and it is then easy to construct inside segment $F_1F_2$ a point $P$ such that $PF_1-PF_2=d$. – Intelligenti pauca Nov 13 '23 at 17:03
  • @Intelligentipauca Thanks. Could you also tell me or provide a link to derive the standard equation of hyperbola from the mess of an equation I got? – Aurelius Nov 15 '23 at 17:47
  • Even if you succeeded at simplifying your formula, you'd obviously get the general equation of a conic but with complicated coefficients. I don't see what you would gain from that. – Intelligenti pauca Nov 15 '23 at 18:25
  • @Intelligentipauca So I was trying to derive equation of random hyperbola from standard definition of a hyerbola. – Aurelius Dec 23 '23 at 22:23
  • Nothing to gain. I was thinking how the equation might have been derived for the first time. This seemed to be the most unique definition of the hyperbola that might be used in its construction. So I tried worked on it. – Aurelius Dec 23 '23 at 22:24

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