wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The point P on the ellipse 4x2+9y2=36 is such that the area of the PF1F2=10 sq. units, where F1,F2 are foci. Then the coordinates of P will be

A
(32,2)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
(32,2)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(32,2)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(32,2)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct options are
A (32,2)
D (32,2)
Given, equation of ellipse
x29+y24=1e=149e=53

Coordinates of foci F1 and F2 are
(±5,0)
Let P(h,k)
Now, the area of
PF1F2=1012×25×|k|=10|k|=2k=2 or 2y=2 or 2

Putting value of y in equation of ellipse
4x2+9(2)2=36x2=92x=32,32

Hence, possible points are
(32,2), (32,2), (32,2), (32,2)

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon