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

Let the point P(α,β) on the ellipse 4x2+3y2=12, in the first quadrant such that the area enclosed by the lines y=x,y=β,x=α and the x axis is maximum, then

A
P(32,1)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
P(32,3)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
eccentric angle of P is π3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
eccentric angle of P is π6
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D eccentric angle of P is π6
Any point P(α,β) on the ellipse x23+y24=1 is (3cosθ,2sinθ)
As P lies in first quadrant, thus
θ(0,π2)


Line PQ:y=2sinθ
Line PR:x=3cosθ
Line OQ:y=x and
Q(2sinθ,2sinθ)

= Area of trapezium PQORP
=12(OR+PQ)PR=12(3cosθ+(3cosθ2sinθ))2sinθ=23cosθsinθ2sin2θ=3sin2θ+cos2θ1=2cos(2θπ3)1
Which is maximum when cos(2θπ3) is maximum
(2θπ3)=0
θ=π6
Hence, point P is (32,1).

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon