AP EAMCET202018 Sep 2020Evening ShiftMathematicsHyperbolaActual
The equation of the transverse axis of hyperbola ((x-3)^2+(y+1)^2=(4 x+3 y)^2 ) is
Options
- A(3 x+4 y=13 )
- B(3 x-4 y=13 )
- C(4 x-3 y=13 )
- D(3 x-4 y=9 )
Correct answer
B. (3 x-4 y=13 )
Step-by-step solution
( aligned (x-3)^2+(y+1)^2 & =(4 x+3 y)^2 (x-3)^2+(y+1)^2 & =25 ( 4 x+3 y 5 )^2 (x-3)^2+(y+1)^2 & =25 ( 4 x+3 y 25 )^2 (x-3)^2+(y+1)^2 & =5 ( 4 x+3 y 5 )^2 aligned ) ( ) It is of the form (S P= ePM ) ( ) Focus (=(3,-1) ) Equation of directrix (=4 x+3 y ) Since, Transverse axis perpendicular to directrix and passing through focus. ( aligned 3 x-4 y+k & =0 3(3)-4(-1)+k & =0 13+k & =0 k & =-13 aligned ) ( ) Required Line is (3 x-4 y-13=0 ) (3 x-4 y=13 ) Hence, option (b) is correct.