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.
Let be any three points on the hyperbola.
We have: .
Find the equation of the altitude from
The slope of is: .
The slope of is: .
The equation of is: .
. . which simplifies to: . .[1]
Similarly, the equation of is: . .[2]
The orthocenter is at: .
Equate [1] and [2] and solve for .
Substitute into [1] and solve for .
Hence, the orthocenter is: .
But note that: .
That is, has coordinates such that:
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.
Just to carry on from where Soroban left off, the x-coordinates of the four points at which the circle meets the hyperbola are given by , or . The product of the four roots is (the constant term). Therefore , and . Those are the negatives of the coordinates that Soroban found for the orthocentre, confirming that this is the point .