Correct Answer - 2
Let `y=mx` be a chord.
Then the points of intersections are given by
`x^(2)(1+m^(2))-x(3+4m)-4=0`
`:. x_(1)+x_(2)=(3+4m)/(1+m^(2))` and `x_(1)x_(2)=(-4)/(1+m^(2))`
Since `(0,0)` divides chord in the ratio `1:4`, we have
`x_(2)= -4x_(1)`
`:. -3x_(1)=(3+4m)/(1+m^(2))` and `4x_(1)^(2)=(-4)/(1+m^(2))`
`:. 9+9m^(2)=9=9+16m^(2)+24m`
i.e., `m=0,-(24)/(7)`
Therefore, the lines are `y=0` and `y+24x=0`