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  1. #1
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    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
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  2. #2
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    Quote Originally Posted by prasum View Post
    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".
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  3. #3
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    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.
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  4. #4
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    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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