KCET2018MathematicsHyperbola
The distance between the foci of a hyperbola is ( 16 ) and its eccentricity is ( 2 ). Its equation is
Options
- A( x²-y²=32 )
- B( x² 4 - y² 9 =1 )
- C( 2 x²-3 y²=7 )
- D( y²-x²=32 )
Correct answer
A. ( x²-y²=32 )
Step-by-step solution
Given that e= 2 and q e=16 So a=4 2 We know that, b²=a² (e²-1 ) So, b²=32(2-1)=32 General equation of hyberbola is given by, x² a² - y² b² =1 Therefore, x²-y²=32