If the lines joining the origin and the point of intersection of the curves $ax^2+2hxy+by^2+2gx=0$ and $a_1x^2+2h_1xy+b_1y^2+2g_1x=0$ are mutually perpendicular then prove that $g (a_1+b_1)=g_1 (a+b) $.
Finding points of intersection seems to tedious.Any better method ?
HINT:
The equation of pair straight lines passing through the intersection of
$$ax^2+2hxy+by^2+2gx=0,a_1x^2+2h_1xy+b_1y^2+2g_1x=0$$
can be written as
$$ax^2+2hxy+by^2+2gx+K(a_1x^2+2h_1xy+b_1y^2+2g_1x)=0$$
$$x^2(a+Ka_1)+xy(\cdots)+y^2(\cdots)+x(\dots)=0\ \ \ \ (1)$$
Now the equation of pair straight lines passing through the origin can be written as $$(y-mx)\left(y+\dfrac1mx\right)=0\iff ?=0\ \ \ \ (2)$$
We need $(1),(2)$ to be identical