Most Important Selected Qs for JEE AdvancedMathematicsHyperbola
Paragraph: The difference between the second degree curve and pair of asymptotes is constant. If second degree curve represented by a hyperbola S=0 , then the equation of its asymptotes is S+ =0 where is constant. Which will be a pair of straight lines, then we get . Then equation of asymptotes is A S+ =0 and if equation of conjugate hyperbola of S represented by S₁ , then A is the arithmetic mean of S and S₁ . Quest
Options
- Ax-y-5=0 and x+y+1=0
- Bx-y=0 and x+y+5=0
- Cx+y-5=0 and x-y-1=0
- Dx+y-1=0 and x-y-5=0
Correct answer
C. x+y-5=0 and x-y-1=0
Step-by-step solution
The transverse axis is the bisector of the angle between asymptotes containing the origin and the conjugate axis is the other bisector. The bisectors of the angle between asymptotes are aligned & (3 x-4 y-1) 5 = (4 x-3 y-6) 5 & (3 x-4 y-1)= (4 x-3 y-6) & x+y-5=0 and x-y-1=0 aligned Hence, transverse axis and conjugate axis are x+y-5=0 and x-y-1=0