q 1 find the locus of the feet of the perpendicular drawn from the point (b,0) on tangents to the - Maths - Conic Sections

Question

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

Ajanta Trivedi · Accepted Answer

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=-1m⇒m=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

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.

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.