If the pair of lines `ax^2+2hxy+by^2= 0 (h^2 > ab)` forms an equilateral triangle with the line `lx + my + n=0` then `(a+3b)(3a+b)=`

Options

A. `H^(2)`
B. `-H^(2)`
C. `2H^(2)`
D. `4H^(2)`

Answer

D

Explanation

Clearly, angle between the lines given by `Ax^(2)+2Hxy+By^(2)=0` must be of `60^(@)`. Therefore,
`tan60^(@)=(2sqrt(H^(2)-AB))/(A+B)`
`rArr" "3(A+B)^(2)=4(H^(2)-AB)rArr(3A+B)(A+3B)=4H^(2)`