CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Let chords of the circle x2+y2=a2 touch the hyperbola x2a2y2b2=1. Then their middle points lie on the curve

A
(x2+y2)2=a2x2b2y2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
(x2+y2)=a2x2b2y2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(x2y2)2=a2x2b2y2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(x2y2)=a2x2b2y2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A (x2+y2)2=a2x2b2y2
The equation of the chord of the circle x2+y2=a2 whose middle point is (α,β) is
xα+yβ=α2+β2y=αβx+α2+β2β
Now the above equation is the tangent to the hyperbola
using the tangency condition
m=αβ,c2=a2m2b2(α2+β2β)2=a2α2β2b2
(α2+β2)2=a2α2b2β2
Hence, the locus of the middle point is
(x2+y2)2=a2x2b2y2

flag
Suggest Corrections
thumbs-up
3
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon