I have have come across an unusual co-ordinate geometry question:
Line $x^2+y^2+axy=0, a \in \Re$ cuts an ellipse $x^2+4y^2=4$ at two points A and B. Find the locus of points of intersection of the tangents drawn at A and B.
In here, a pair of family of lines passing through origin has been referred to as a line! Also a pair of lines would cut the ellipse at four points and not at two!!
I may come back. Your comment/suggestion/solution is welcome.
EDIT: The suggested solution in the nondescript reference that I came across asks to take the point $(h,k)$ and write the equation of the chord of contact of the ellipse as $hx+4ky=4$, next homogenize it with the ellipse as $$x^2+4y^2=4\left(\frac{hx+4ky}{4}\right)^2$$ the combined equation of OA and OB as $$(1-h^2/4)x^2+(4-4k^2)y^2-2hkxy=0.....(*)$$ and compare it with another combined equation of OA and OB namely $x^2+y^2+axy=0$ to get a connection between $h$ and $k$ to get the desired locus one gets: $1-h^2/4=4-4k^2$. Hence the required locus is $$x^2-16y^2+12=0.$$
Interestingly, the pole of a line passing through the center of a circle or conic either doesn't exist or it is found at infinity. In other words, the tangents at the end points of these lines are parallel. So as also pointed by @Kurt G. we can ignore the other two ends of pair of lines and work out the locus of point of intersection of tangents put of points A and B. Let the equation of lines be $y=mx$ and $y=x/m$.
The points of intersection of these lines with the given ellipse $x^2+4y^2=4$ are $$A\left(\frac{1}{2\sqrt{1+4m^2}},\frac{2m}{\sqrt{1+4m^2}}\right);\quad B\left(\frac{m}{2\sqrt{4+m^2}},\frac{2}{\sqrt{4+m^2}}\right).$$ Equations of tangents on the ellipse at these point are $$x+4my=2\sqrt{1+4m^2}; \quad mx+4y=2\sqrt{4+m^2}$$ The point of intersection C of these two tangents is found to be $$C=(x,y)= \left(2\frac{\sqrt{1+4m^2}-m\sqrt{4+m^2}}{2(m^2-1)},\frac{\sqrt{4+m^2}-m\sqrt{1+4m^2}}{2(m^2-1)}\right).$$ Thankfully, 'Eliminate' command helps in eliminaion the eliminant is the required locus: $$x^2-16y^2+12=0,$$ which is exactly the same as obtained in the EDIT part of my problem proposal in the post.