Kepler gave the following construction for a hyperbola with foci at A and B and

Question

Kepler gave the following construction for a hyperbola with foci at A and B and with one vertex at C: Let pins be placed at A and B. To A let a thread with length AC be tied and to B a thread with length BC. Let each thread be lengthened by an amount equal to itself. Then grasp the two threads together with one hand (starting at C) and little by little move away from C, paying out the two threads. With the other hand, draw the path of the join of the two threads at the fingers. Show that the path is a hyperbola.

Explanation / Answer

We know that a hyperbola is locus of a point that moves such that the difference of dstances between two fixed points is a constant. Now in the construction explained above, initially the difference is length of string BC - length of string AC which is constant. (say a-b) The two strings are grasped at C with still some part of the strings in loosened position. Now that the strings are moved away from C slowly, each string will get loosened by same amount say 'r'. So at any new position of the fingers, the length of the two strings will be 'a+r' and ' b+r' respectively. The difference in the distance of the point from foci is nothing but the difference between the lengths of strings at that position. i.e ((a+r) - (b+r)) = (a-b) , a constant. Hence the locus is a hyperbola PB - PA = a-b

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