The slope(s) of the common tangent(s) to the two hyperbolas x2a2−y2b2=1 and y2a2−x2b2=1 is/are
A
1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
−1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
−12
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C−1 Let the slope of tangent is m, then
Tangent to x2a2−y2b2=1 is y=mx±√a2m2−b2⋯(1)
The tangent to other hyperbola y2a2−x2b2=1 is y=mx±√a2−b2m2⋯(2) ∵ both are same equations ∴a2m2−b2=a2−b2m2 ⇒(a2+b2)m2=a2+b2 ⇒m=±1