I am trying to find a conic curve that passes through $[0:1:2]$ and intersects the conic curve $F =x_1^2 + 4x_1x_2 + 4x_0x_2 -x_0^2$ at the point $[1:1:0]$ with multiplicity $4$. Since I am given $5$ points (counted with multiplicity) there should be a unique conic that fulfils the requested properties.
Projective conic curves are of the form: $Ax_1^2 + Bx_1x_2 + Cx_2^2 + Dx_1x_0 + Ex_2x_0 + Fx_0^2$.
Since it has to go through the points $[0:1:2]$ and $[1:1:0]$ I get $A + 2B + 4C = 0$ and $A+D+F = 0$.
However, from here I don't know how I can proceed and how I can use the fact, that the intersection multiplicity in $[1:1:0]$ has to be $4$.