Question

Solution

Correct option is

4 (x2 – y2) = a2 – b2

Let C (hk) be the centre of the circle passing through the end points of the rod AB and PQ of length a and b respectively, CL and CM be perpendiculars from C on AB and PQ respectively. (Fig.)  PM = (1/2) PQ = b/2

and CA = CP (radii of the same circle)  So that locus of (hk) is 4(x2 – y2) = a2 – b2

SIMILAR QUESTIONS

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Q10

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