1. ## hyperbola

If P(x1,y1), Q(x2,y2) R(x3,y3) S(x4,y4) are four concyclic points on the rectangular hyperbola xy=c^2 then the coordinates of orthocenter of Triangle PQR is

2. Originally Posted by prasum
If P(x1,y1), Q(x2,y2) R(x3,y3) S(x4,y4) are four concyclic points on the rectangular hyperbola xy=c^2 then the coordinates of orthocenter of Triangle PQR is
The answer seems to be $(-x_4, -y_4)$. In other words, the orthocentre is the point on the hyperbola opposite the point S. For further information on this, do a search for "Feuerbach hyperbola".

3. Hello, prasum!

I have not solved this exact problem,
. . but I learned a surprising fact about hyperbolas.
And, as you'll see, I skipped all the tedious algebra.

$\text{If }P(x_1,y_1), Q(x_2,y_2), R(x_3,y_3), S(x_4,y_4)\text{ are four concyclic points}$
$\text{on the rectangular hyperbola }xy\,=\,c^2,$
$\text{then the coordinates of orthocenter of }\Delta PQR \text{ is }\_\_.$

Let $P,Q,R$ be any three points on the hyperbola.

We have: . $P\left(x_1,\,\dfrac{c^2}{x_1}\right),\;Q\left(x_2, \:\dfrac{c^2}{x_2}\right),\;R\left(x_3,\:\dfrac{c^ 2}{x_3}\right)$

Find the equation of the altitude from $\,P.$

The slope of $QR$ is: . $m \;=\;\dfrac{\frac{c^2}{x_2} - \frac{c^2}{x_3}}{x_2-x_3} \;=\;\dfrac{c^2(x_3-x_2)}{x_2x_3(x_2-x_3)} \;=\;-\dfrac{c^2}{x_2x_3}$

The slope of $\text{Alt}_P$ is: . $m_P \:=\:\dfrac{x_2x_3}{c^2}$

The equation of $\text{Alt}_P$ is: . $y - \dfrac{c^2}{x_1} \;=\;\dfrac{x_2x_3}{c^2}(x-x_1)$

. . which simplifies to: . $y \;=\;\dfrac{x_2x_3}{c^2}x + \dfrac{c^4 - x_1^2x_2x_3}{c^2x_1}$ .[1]

Similarly, the equation of $\text{Alt}_Q$ is: . $y \;=\;\dfrac{x_1x_3}{c^2}x + \dfrac{c^4 - x_1x_2^2x_3}{c^2x_2}$ .[2]

The orthocenter is at: . $\text{Alt}_P \cap \text{Alt}_Q$

Equate [1] and [2] and solve for $\,x\!:$ . $x \;=\;-\dfrac{c^4}{x_1x_2x_3}$

Substitute into [1] and solve for $\,y\!:$ . $y \;=\;-\dfrac{x_1x_2x_3}{c^2}$

Hence, the orthocenter $\,H$ is: . $\left(-\dfrac{c^4}{x_1x_2x_3},\;-\dfrac{x_2x_2x_3}{c^2}\right)$

But note that: . $\left(-\dfrac{c^4}{x_1x_2x_3}\right)\left(-\dfrac{x_1x_2x_3}{c^2}\right) \;=\;c^2$

That is, $\,H$ has coordinates $(x,y)$ such that: $xy \:=\:c^2$

Therefore, the orthocenter of any triangle on a hyperbola lies on the hyperbola.

No doubt this was established centuries ago, but it was a surprise to me.

4. Just to carry on from where Soroban left off, the x-coordinates of the four points at which the circle $x^2+y^2+2gx+2fy+k=0$ meets the hyperbola $xy=c^2$ are given by $x^2+\bigl(\frac{c^2}x\bigr)^2+2gx+2f\bigl(\frac{c^ 2}x\bigr)+k=0$, or $x^4 + 2gx^3 + kx^2 + 2fc^2x + c^4 = 0$. The product of the four roots is $x_1x_2x_3x_4 = c^4$ (the constant term). Therefore $x_4 = \frac{c^4}{x_1x_2x_3}$, and $y_4 = \frac{c^2}{x_4} = \frac{x_1x_2x_3}{c^2}$. Those are the negatives of the coordinates that Soroban found for the orthocentre, confirming that this is the point $(-x_4,-y_4)$.

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# If P,Q,R,S are four concyclic points on the rectangular hyperbola xy=c^2 ,the coordinates of the orthocentre of triangle PQR are

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