1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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)

A

2 sq. units

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

3 sq. units

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C

23 sq. units

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

4 sq. units

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B √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

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Perpendicular Distance of a Point from a Plane
MATHEMATICS
Watch in App
Join BYJU'S Learning Program