If the acute angle between the lines `ax^2+2hxy +by^2=0` is `60^@` then show that `(a+3b)(3a+b)= 4h^2`

If the acute angle between the lines x^(2)-4hxy+3y^(2)=0 is 60^(@) then h=

If the acute angle between the lines x^(2)-4hxy+3y^(2)=0 is 60^(@) then h=

If acute angle between lines x^(2)-2hxy+y^(2)=0 is 60^(@) then h=

If angle between lines ax^(2)+2hxy+by^(2)=0 is (pi)/4 then 2h=

If acute angle between lines ax^(2)+2hxy+by^(2)=0 is (pi)/4 , then 4h^(2)=

If theta is the acute angle between the lines given by ax^(2) + 2hxy + by^(2) = 0 then prove that tan theta = |(2 sqrt (h^(2) - ab))/(a + b)| . Hence find acute angle between the lines 2x^(2) + 7xy + 3y^(2) = 0

If acute angle between lines 3x^(2)-4xy+by^(2)=0 is cot^(-1)2, then b=

If the acute angle between the lines ax^(2)+2hxy+by^(2)=0 is congruent to the acute angle between the lines 3x^(2)-7xy+4y^(2)=0 , then

If the angle between the lines represented by ax^(2) + 2hxy + by^(2) = 0 is equal to the angle between the lines 2x^(2) - 5xy + 3y^(2) = 0 , then show that 100(h^(2)-ab) = (a+b)^(2) .

Show that if one of angle betwwen pair of straight lines ax^(2)+2hxy+by^(2)=0 is 60^(@) then (a+3b)(3a+b)=4h^(2)