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A circle cuts a rectangualar hyperbola xy = c2 at A, B, C and D. If H is the orthocentre of ΔABC, then show that H and D are the extremities of a diameter of the curve.

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If (ctr ,c/tr) (where r = 1, 2, 3 and 4) are points A, B, C and D, respectively, the orthocentre H of triangle ABC is

which also lies on the curve. Now, since t1t2t3t4 , we have

t4 = 1/t1t2t3

so that

and hence (0, 0) is the midpoint of HD.

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