AP EAMCET202022 Sep 2020Evening ShiftMathematicsHyperbolaActual
The equation of the hyperbola with focus (1,2), e= 3 and directrix 2 x+y=1 is given by
Options
- A2 y^2-12 x y-7 x^2+2 x-14 y+22=0
- B2 y^2+12 x y+7 x^2-2 x+14 y-22=0
- C2 y^2-12 x y-7 x^2-2 x-14 y-22=0
- D2 y^2+12 x y+7 x^2+2 x+14 y+22=0
Correct answer
A. 2 y^2-12 x y-7 x^2+2 x-14 y+22=0
Step-by-step solution
Given, Focus (S)=(1,2) Eccentricity (e)= 3 Equation of Directrix is 2 x+y=1 Required equation of hyperbola is S P=e PM (x-1)^2+(y-2)^2 = 3 |2 x+y-1| 2^2+1^2 Squaring on both sides, aligned & (x-1)^2+(y-2)^2= 3 5 (2 x+y-1)^2 & x^2+1-2 x+y^2+4-4 y = & 3 5 (4 x^2+y^2+1+4 x y-2 y-4 x ) & 5 (x^2+y^2-2 x-4 y+5 ) = & 3 (4 x^2+y^2+4 x y-4 x-2 y+1 ) & 5 x^2+5 y^2-10 x-20 y+25 & =12 x^2+3 y^2+12 x y-12 x-6 y+3 & 5 x^2+5 y^2-10 x-20 y+25-12 x^2 & -3 y^2-12 x y+12 x+6 y-3=0 & 2 y^2-7 x^2-12 x y-7 x^2+2 x-14 y+22=0 aligned Henc