The equations $x-y=4$ and $x^2+4xy+y^2=0$ represent the sides of a/an

The normal to the curve $x^2 + 2xy - 3y^2 = 0$ at the point $(1,1)$ intersects the curve again in which quadrant?

Area of the triangle bounded by the lines given by the equations $12 x^2-20 x y+7 y^2=0$ and $x+y-1=0$ is

The lines represented by the equations $23x^2 - 48xy + 3y^2 = 0$ and $2x + 3y + 4 = 0$ form

Let $P$ be the pair of lines represented by $2x^2 - 5xy + 2y^2 + 6x - 3y = 0$. Consider the following independent statements:
$(i)$ $\alpha$ is the $x$-coordinate of the point of intersection of the pair of lines $P$.
(ii) $\beta$ is the slope of one of the lines of $P$ passing through the origin.
(iii) $\gamma$ is the constant term in the equation of the pair of angular bisectors of $P$.
Then,

The numbers $\alpha$ and $\beta$ are such that one of the lines of $2x^2 + \alpha xy + 3y^2 = 0$ coincides with one of the lines of $2x^2 + \beta xy - 3y^2 = 0$. If the two lines other than that common line are perpendicular,then $|\alpha + \beta|$ is equal to