The equation $(x^2 + αy^2 + β) (y^2 – xy – 2x^2) = 0$ always represents
(A) A circle and a pair of straight lines if α = 1 and β = 0
(B) A pair of straight line and an ellipse if α ≠ 1 and α > 0
(C) A pair of hyperbolas if α < 0
(D) A pair of straight lines and a hyperbola if α < 0 and β ≠ 0
My approach $(y^2 – xy – 2x^2)=0$ represent a pair of straight line the lines are $y-2x=0$ & $y+x=0$ , Hence Choice C gets eliminated as pair of straight line option is not there.
$(x^2 + αy^2 + β) = 0$ case, if α = 1 and β = 0 it is a case of a point in lieu of circle hence A is also eliminated
$(x^2 + αy^2 + β) = 0$ case, if α ≠ 1 and α > 0 it can be an ellipse but not sure about the value of "β".
$(x^2 + αy^2 + β) = 0$ case, if β ≠ 0 and α <0 it is a case of hyperbola
My query is regarding choice (B)
For $(x^2 + αy^2 + β) = 0$, if α ≠ 1 and α > 0, β must be negative for it to be an ellipse. So, it is not always true and (B) is not the answer.
For $(x^2 + αy^2 + β) = 0$, sine β ≠ 0 and α <0, it can always be cast in a standard hyperbola form, either with major axis along x or y depending on the sign of β. Therefore, (D) should be the choice.