Question

# Let Z1, Z2, Z3 be three points A, B and P respectively in the argand plane. Let P moves in the plane such that |Z−Z1|+|Z−Z2|=K. Let Z1=–Z2=i Maximum area of the triangle ABP is (if K = 4) 2 sq. units 2√3 sq. units √3 sq. units 4 sq. units

Solution

## The correct option is C √3 sq. units |Z−Z1|+|Z−Z2|=K is the condition of an ellipse. It is a standing ellipse with foci (0, 1) and (0, −1) ⇒be=1  [Comparing it with standard equation of ellipse : x2a2+y2b2=1 with eccentricity, e] ⇒b×√1−a2b2=1 ⇒√b2−a2=1 ⇒b2−a2=1 Now, b=2∵K=4 ⇒a=√3 AB=2  Area of the triangle ABP is maximum when P is at (√3, 0) or (−√3, 0) So, area of the triangle ABP=12×AB×√3=√3 sq. units

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