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

PM and PN are the perpendiculars from any point P on the rectangular hyperbola xy=8 to the asymptotes. If the locus of the mid point of MN is a conic, then the least distance of (1,1) to director circle of the conic is

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

The correct option is D 2 unit


Here, OMPN is rectangle.
P(22t,22t)

Mid point of MN=OP
(h,k)=(2t,2t)
Thus h×k=2t2t=2
Thus locus is the rectangular hyperbola xy=2
The director circle to this rectangular hyperbola is the point circle with center (0,0).
Thus distance between (0,0) and (1,1) is 2 unit


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Parametric Representation-Hyperbola
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon