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

The centroid of the triangle formed by the pair of straight lines $12x^2 - 20xy + 7y^2 = 0$ and the line $2x - 3y + 4 = 0$ is:

One line of the pair of lines $x^2+xy-2y^2=0$ is perpendicular to one line of the pair of lines $3y^2-5xy-2x^2=0$. If the combined equation of the two lines other than those two perpendicular lines is $ax^2+2hxy+by^2=0$,then $a+2h+b=$

If the pair of straight lines given by $Ax^2+2Hxy+By^2=0$ $(H^2>AB)$ forms an equilateral triangle with the line $ax+by+c=0$,then $(A+3B)(3A+B)$ is equal to:

$2 x^2-3 x y-2 y^2=0$ represents two lines $L_1$ and $L_2$. $2 x^2-3 x y-2 y^2-x+7 y-3=0$ represents another two lines $L_3$ and $L_4$. Let $A$ be the point of intersection of lines $L_1$ and $L_3$,and $B$ be the point of intersection of lines $L_2$ and $L_4$. The area of the triangle formed by lines $AB$,$L_3$,and $L_4$ is

If the slopes of both the lines given by $x^2 + 2hxy + 6y^2 = 0$ are positive and the angle between these lines is $\operatorname{Tan}^{-1}\left(\frac{1}{7}\right)$,then the product of the perpendiculars from the point $(1, 1)$ to these lines is: