If the represented by the equation `3y^2-x^2+2sqrt(3)x-3=0` are rotated about the point `(sqrt(3),0)` through an angle of `15^0` , on in clockwise direction and the other in anticlockwise direction, so that they become perpendicular, then the equation of the pair of lines in the new position is (a)`y^2-x^2+2sqrt(3)x+3=0` (b)`y^2-x^2+2sqrt(3)x-3=0` (c)`y^2-x^2-2sqrt(3)x+3=0` (d) `y^2-x^2+3=0`

Options

A. `y^(2)-x^(2)+2sqrt(3)x+3=0`
B. `y^(2)-x^(2)+2sqrt(3)x-3=0`
C. `y^(2)-x^(2)-2sqrt(3)x+3=0`
D. `y^(2)-x^(2)+3=0`

Answer

2

Explanation

The given equation of the pair of straight lines can be rewritten as
`(sqrt(3)y-x+sqrt(3))(sqrt(3)y+x-sqrt(3))=0`
Their separate equation are
`sqrt(3)y-x+sqrt(3)=0andsqrt(3)y+x-sqrt(3)=0`
or `y=(1)/(sqrt(3))x-1andy=-(1)/(sqrt(3))x+1`
or `y=(tan30^(@))x-1andy=(tan150^(@))x+1`

After rotation through an angle of `15^(@)` as given in the question, the lines are
`(y-0)=tan45^(@)(x-sqrt(3))and(y-0)=tan135^(@)(x-sqrt(3))`
or `y=x-sqrt(3)and y=-x+sqrt(3)`
Their combined equation is
`(y-x+sqrt(3))(y+x-sqrt(3))=0ory^(2)-x^(2)+2sqrt(3)x-3=0`