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Question

Let m1,m2 are slopes of two tangent that are drawn from (2,3) to the parabola y2=4x and α is the harmonic mean of m1&m2 if β is the value of (1+tan23)(1+tan22) ,then 32α+β is

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Solution

(1+tan23o)(1+tan(45o23o))
=(1+tan23o)(1+tan45otan23o1+tan45otan23o)
(tan(AB)=tanAtanB1+tanAtanB)
=(1+tan23o)(1+1tan23o1+tan23o) (tan45o=1)
=(1+tan23o)(1+tan23o+1tan23o)(1+tan230)
=2=β
If m1 & m2 are slopes of tangents,
we know they intersect
at (a(m1+m2),am1m2) where
y2=4ax is eqn of parabola
Here, a=1
m1+m2=2,m1m2=3
Harmonic mean of m1 & m2
=21m1+1m2=2m2+m1m1m2
=2m1m2m1+m2=2×32
=3=α
32α+β=3×32+2=132

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