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Let \mathrm{P} and \mathrm{Q} are any two points on the circle \mathrm{\mathrm{x}^{2}+\mathrm{y}^{2}=4} such that \mathrm{\mathrm{PQ}} is a diameter. If \mathrm{L}_{1} and \mathrm{L}_{2} are the lengths of perpendiculars from \mathrm{P} and \mathrm{Q} on \mathrm{x}+\mathrm{y}=1,then the maximum value of \mathrm{L}_{1} \mathrm{~L}_{2} is

Option: 1

1 / 2


Option: 2

1


Option: 3

7 / 2


Option: 4

2


Answers (1)

best_answer

Let  \mathrm{P} \equiv(2 \cos \theta, 2 \sin \theta)
\& \mathrm{Q} \equiv(-2 \cos \theta,-2 \sin \theta)

\therefore \mathrm{L}_{1} \mathrm{~L}_{2}
=\frac{|2 \cos \theta+2 \sin \theta-1| \cdot|-2 \cos \theta-2 \sin \theta-1|}{2}
=\frac{\left|4(\cos \theta+\sin \theta)^{2}-1\right|}{2} \leq \frac{7}{2}

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