How to find point of intersection with conic section and tangent dropped from a point not on a conic Question: We have conic section $-12x^2 + 28xy+4x-9y^2-8y=0$ and a point not on a conic $(2/5,1/5)$, how to find an intersection point with tangent dropped from $(2/5,1/5)$ to a conic section?
My ideas: to use polar points and Homogeneous coordinates

The given equation of the conic is
$ - 12 x^2 + 28 xy - 9 y^2 + 4 x - 8 y = 0 \hspace{15pt}(1) $
And the given point is $(\dfrac{2}{5}, \dfrac{1}{5})$
Using implicit differentiation, with respect to $x$,
$ - 24 x + 28 (y + x y') - 18 y y' + 4 - 8 y' = 0 $
Therefore,
$ y' (28 x - 18 y - 8) = 24 x - 28 y - 4 $
i.e.
$ y' = \dfrac{ 24 x - 28 y - 4 }{28 x - 18 y - 8} = \dfrac{12 x - 14 y - 2}{14 x - 9 y - 4 } $
Now $y'$ is the slope of the tangent line, which is also given by
$ y' = \text{Slope} = \dfrac{ \dfrac{1}{5} - y }{\dfrac{2}{5} - x} $
Therefore, by cross multiplication,
$ ( 12 x - 14 y  - 2 ) (\dfrac{2}{5} - x) = (14 x - 9 y - 4) (\dfrac{1}{5} - y) $
Multiply through by $5$
$ (12 x - 14 y - 2) (2 - 5 x) = (14 x - 9 y - 4) ( 1 - 5 y)$
Expand
$ 24 x - 60 x^2 - 28 y+ 70 x y - 4 + 10 x = 14 x - 9 y - 4 - 70 x y + 45 y^2 + 20 y$
And this reduces to
$ -60 x^2 + 140 x y - 45 y^2 + 20 x -39 y = 0 \hspace{15pt} (2)$
Now we have to solve the system of equations $(1),(2)$, and the solutions are
$(0, 0) $ and $(\dfrac{1}{3} , 0 ) $
