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

The line y = mx + c becomes a tangent to the hyperbola x2a2 − y2b2 = 1,then the value of c is


A

± a2m2+b2

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

± a2m2b2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C

± a2m2+a2

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

± a2m2a2

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B

± a2m2b2


For questions like these in which number of points of intersection are concerned we first solve the

equation .we get quadratic on any one variable .we then put determine as 0,+ve.or -ve according

to 1,2 and 0 number of solutions respectively .Solving equations of hyperbola and line

x2a2 (mx+c)2b2 = 1

x2(a2m2b2)+ × (2a2mc)+a2(a2+b2)=0.

Given that line touches the hyperbola No.of solutions =1.

Determinant = 0.

i.e.,4a2m2c24a2(c2+b2)(a2m2b2)=0.

e2=a2m2b2

c = ±a2m2b2 (option b)

Hence we get equations of line as,

y = mx ± a2m2 b2

Also for secant

c2 > a2m2b2

For line not interestingtouching hyperbola at all.

c2 < a2m2b2


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Lines and Points
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon