NTA Abhyas JEE Main2020MathematicsHyperbolaPractice
The equation of a hyperbola is x 2 a 2 - y 2 b 2 = 1 . If P 2 , 5 is a point from which perpendicular tangents can be drawn to the hyperbola and distance between both the foci of the hyperbola is 10 , then its eccentricity is
Options
- A3 2
- B5 4
- C16 9
- D4 3
Correct answer
B. 5 4
Step-by-step solution
∵ P lies on director circle x 2 + y 2 = a 2 - b 2 ∴ a 2 - b 2 = 7 …(i) Also, 2 a e = 10 ⇒ a e = 5 …(ii) Now, e 2 = a 2 + b 2 a 2 ⇒ a 2 + b 2 = 25 …(iii) From (i) & (iii), we get, a 2 = 16 , b 2 = 9 e = 25 16 = 5 4