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Question

A circle circumscribing the triangle formed by three co-normal points passes through the vertex of the parabola y2=4ax where (h,k) is the point from where three concurrent normals are drawn. The circle has the equation

A
2(x2+y2)(h+2a)+ky=0
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B
2(x2+y2)2(h+2a)+ky=0
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C
2(x2+y2)2(h+a)+ky=0
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D
none of these
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Solution

The correct option is B 2(x2+y2)2(h+2a)+ky=0
The equation of the normal at (am2,2am) is
y+mx=2am+am3
It passes through (h,k).
am3+m(2ah)k=0
The roots are m1,m2 and m3
m1=0,m1m2=2aha
m1m2m3=ka
Let equation of circle be x2+y2+2gx+2fy+C=0
It passes through (am2,2am)
a2m4+4a2m2+2agm2+4afm+C=0
a2m4+m2(4a2+2ag)+4afm+C=0
Its roots are m1,m2,m3 and m4
m1+m2+m3+m4=0
m1+m2+m3=0
m4=0 the circle passes (0,0)
m1m2+m2m3+m3m4+m4m1+m1m3+m2m4
=4a2+2aga2
2aha=4a2+2aga2
2ah=4a+2g
g=h2a2
m1m2m3+m2m3m4+m3m4m1+m4m1m2=4afa2
ka=4afa2
f=k4
The equation of circle is,
x2+y2(h+2a)x+k2y=0
2(x2+y2)2(h+2a)+ky=0

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