KCET2023MathematicsHyperbola
The distance between the foci of a hyperbola is 16 and its eccentricity is 2 . Its equation is
Options
- Ax^2 4 - y^2 9 =1
- B2 x^2-3 y^2=7
- Cy^2-x^2=32
- Dx^2-y^2=32
Correct answer
D. x^2-y^2=32
Step-by-step solution
Given, distance between the foci =16 and aligned e & = 2 2 a e & =16 a e & =8 aligned a= 8 2 =4 2 So, aligned & b^2=a^2 (e^2-1 ) & b^2=32(2-1)=32 aligned Equation of hyperbola is aligned & x^2 a^2 - y^2 b^2 =1 & x^2 32 - y^2 32 =1 & x^2-y^2=32 aligned