AP EAMCET201822 Apr 2018Morning ShiftMathematicsHyperbolaActual
The equation of the hyperbola whose asymptotes are the lines 3 x+4 y-2=0 , 2 x+y+1=0 and which passes through the point (1,1) is
Options
- A6 x^2+11 x y+4 y^2-30 x+2 y+7=0
- B6 x^2+11 x y+4 y^2-x+2 y-22=0
- C6 x^2+11 x y+4 y^2-x+2 y+22=0
- D6 x^2+11 x y+4 y^2-3 x-7 y-11=0
Correct answer
B. 6 x^2+11 x y+4 y^2-x+2 y-22=0
Step-by-step solution
Equation to the asymptotes are given as aligned 3 x+4 y-2 & =0 2 x+y+1 & =0 aligned and Eqs.(i) and (ii) may be given by (3 x+4 y-2)(2 x+y+1)=0 As, the equation to the hyperbola will differ from Eq. (iii) only by a constant, it may be given by (3 x+4 y-2(2 x+y+1)= (where is a constant) (1,1) lies on the curve given by Eq. (iv), we have aligned & & (3+4-2)(2+1+1) & = & & (5)(4)= & =20 aligned Hence, the equation of the hyperbola will be aligned & (3 x+4 y-2)(2 x+y+1)=20 & 6 x^2+3 x y+3 x+8 x y+4 y^2 & +4 y-4 x-2 y-2