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Question

A parabola touches two given straight lines; if its axis pass through the point (h, k), the given lines being the axes of coordinates, prove that the locus of the focus is the curve
x2y2hx+ky=0.

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Solution

When axes is inclined focus is given by

ax=by=x2+2xycosω+y2a=x2+y2x,b=x2+y2y.......(i)

The abssica of focus is x=ab2a2+2abcosω+b2..........(ii)

Axis of parabola is given by

aybx=ab(a2b2)a2+2abcosω+b2

It passes through (h,k)

akbh=ab(a2b2)a2+2abcosω+b2akbh=ab2a2+2abcosω+b2×a2b2b

Substituting (ii)

akbh=x(a2b2)b

Substituting a and b from (i)

(x2+y2x)k(x2+y2y)h=x(x2+y2x)2(x2+y2y)2(x2+y2y)kxhy=xy(1x21y2)x2y2hx+ky=0

Hence proved.


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