Point of intersection of a given straight line with 2 degree curve. I am currently solving a question in which there is equation as $5x^2+12xy-6y^2+4x-2y+3=0$ and a straight line $x+ky=1$. In the question it mentions that the straight lines joining points of intersection of given 2 equations and origin are equally inclined. Please help me find the points of intersection or help me with some method to solve this.
 A: Hint:
Replace $x=1-ky$ to form a Quadratic Equation $(1)$ in $y$
Now if the roots are $y_1,y_2,$ the points of intersections will be  $(1-ky_1,y_1);(1-ky_2,y_2)$
By "equally inclined" I understand the sum of the two gradients is zero
$$\implies\dfrac{y_1-0}{1-ky_1-0}=-\dfrac{y_2-0}{1-ky_2-0}$$
$$\implies y_1+y_2=2ky_1y_2$$
Replace the values of $y_1+y_2,y_1y_2$ in terms of $k$ from $(1)$ 
A: It is a verbal clue. If slope of two straight lines passing through origin is same then they are a double straight line set of the form:
$$ (y-mx)^2=0 $$
where $m$ can be left that way as an arbitrary slope constant, forgetting $k$.
Accordingly plug in $ y= mx $ into the given equation of a tilted hyperbola and solve for points of intersection from the resulting quadratic equation.
A: By homogenisation the equation to the pair of lines is 
$$ 5x^2+12xy-6y^2+(4x-2y)(x+ky)+3(x+ky)^2=0$$
If the lines are equally inclined they will be of the form $y^2-m^2x^2=0$ i.e. the $xy$ term vanishes.
In the above equation, coefficient of $xy$ is $10(1+k)=0 \Rightarrow \boxed{k=-1}$
