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

# If the tangents drawn to the hyperbola 4y2=x2+1 intersect the co-ordinate axes at the distinct points A and B, then the locus of the mid point of AB is :

A
x24y2+16x2y2=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
x24y216x2y2=0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
4x2y2+16x2y2=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
4x2y216x2y2=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B x2−4y2−16x2y2=0Any point on the hyperbola will be, (tanθ,secθ2) Equation of tangent, T=0 4ysecθ2=xtanθ+1⇒2ysecθ=xtanθ+1 Given that tangents drawn to the hyperbola intersect the co-ordinate axes at the distinct points A and B Thus the coordinates of A y=0,x=−cotθ And the coordinates of B x=0,y=cosθ2 Let the midpoint of AB=(h,k) (h,k)=⎛⎜ ⎜ ⎜⎝−cotθ+02,0+cosθ22⎞⎟ ⎟ ⎟⎠⇒−12h=tanθ,14k=secθ⇒(14k)2−(−12h)2=1⇒4h2−16k2=64k2h2 Hence the locus of the midpoint will be, x2−4y2−16x2y2=0

Suggest Corrections
2
Join BYJU'S Learning Program
Related Videos
Lines and Points
MATHEMATICS
Watch in App
Join BYJU'S Learning Program