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If two distinct chords of a parabola \mathrm{y^2=4 a x}, passing through \mathrm{(a, 2 a)} are bisected by the line \mathrm{x+y=1}, then maximum pouble integral value that can be assigned to the length of latus rectum is

Option: 1

3


Option: 2

0


Option: 3

6


Option: 4

2


Answers (1)

best_answer

Any point on the line \mathrm{x+y=1} can be taken as \mathrm{(t, 1-t)}. Equation of chord with \mathrm{(t, 1-t)} as mid point is \mathrm{y(1-t)-2 a(x+t)=(1-t)^2-4 a t}

It passes through \mathrm{(a, 2 a) \therefore t^2-2 t+2 a^2-2 a+1=0} This should have two distinct real roots so \mathrm{a^2-a<0\Rightarrow 0<a<1 \Rightarrow}Length of latus rectum \mathrm{ <4}

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Sayak

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