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 $\displaystyle (-x_4, -y_4)$. 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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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# concyclic points on rectangular hyperbola

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