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Question

The condition that the two tangents to the parabola y2=4ax become normal to the circle x2+y2−2ax−2by+c=0 is given by

A
a2>4b2
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B
b2>2a2
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C
a2>2b2
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D
b2>4a2
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Solution

The correct option is D b2>4a2
we have
x2+y22ax2by+c=0
center will be {(2a)2,(2b)2}=(a,b)
Pain of tangents will par through center of circle or
in other wads, per of length must originate from
center of circle
we have
y2=4acx
S:y24ax=0
(x1,y1)=(a,b)
S1y214ax1=0
T:yy14a(x1+x2)=0
T=yy12a(xx1)=0
Equation of pan of tangents is :
SS1=T2
(y24ax)(y214ax1)={yy12a(x+x1)}2
(y24ax)(b24a.a)={yb2a(x+a)}2 on substituent (x1,y2)(a,b)
(y24ax)(b24a2)={by2a(x+a)}2
from here, b2 should be grater then 4a2
then b2>4a2


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