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Question

If the lines (y−b)=m1(x+a) and (y−b)=m2(x+a) are two tangents to the parabola y2=4ax then

A
m1+m2=0
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B
m1m2=1
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C
m1m2=1
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D
m1+m2=1
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Solution

The correct option is A m1m2=1
Condition for tangency is c=am
Hence
(am1b)m1=a
(am2b)m2=a
On comparing we get
am12bm1=am22bm2
a(m21m22)=b(m1m2)

a(m1m2)(m1+m2)=b(m1m2)

m1+m2=ba
Now (m1ba)m1=1
m1m2=1

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