q.1 find the locus of the feet of the perpendicular drawn from the point (b,0) on tangents to the circle x2+y2 =a2.

the equation of the tangent to the circle x2+y2=a2 ......(1) is y=mx±a1+m2 .......(2)
let P(h,k) be the locus of the feet of the perpendicular drawn from the point A(b,0).
therefore PA is perpendicular to the tangents 
slope of PA = - 1/m
k-0h-b=-1mm=b-hk.......(3)

and point (h,k) must lie on the tangent (2), therefore
k=mh±a1+m2from eq(3)k=(b-h)k.h±a.1+b-hk2k-(b-h).hk=±a1+b-hk2k2-bh+h2k=±ak2+(b-h)2k2on squaring both sides , we have:(k2+h2-bh)2k2=a2.[k2+(b-h)2k2](k2+h2-bh)2=a2.[k2+(b-h)2]
now for locus replace h by x and k by y:
y2-bx+x22=a2.y2+(b-x)2

hope this helps you

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