wiz-icon
MyQuestionIcon
MyQuestionIcon
8
You visited us 8 times! Enjoying our articles? Unlock Full Access!
Question

Locus of the point which divides double ordinates of the ellipse x2a2+y2b2=1,a>b in the ratio 1:2 internally is

A
x2a2+9y2b2=1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
x2a2+9y2b2=19
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
9x2a2+9y2b2=1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
x2a2+y2b2=a2b2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A x2a2+9y2b2=1
Given that : x2a2+y2b2=1
Let end points of double ordinates be P(α,β) and Q(α,β)
Let the point which divides PQ in 1:2 is R(h,k)
Then, by internal section formula
h=α
And
k=1×β2×β3 or k=2×β1×β3β=±3k

(α,β) lies on ellipse
h2a2+(±3k)2b2=1

Hence, the locus of R is
x2a2+9y2b2=1

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