I want to solve the following problem:
We know that the irreducible quartic $\mathcal{C}=\mathrm{V}(Y^{2}+X^{2}+X^{2}Y^{2}-2XY(X+Y+1))$ has three singular points (which are three cusps) in $P=(0:0:1)$, $Q=(0:1:0)$ and $R=(1:0:0)$ with respective tangent lines $x=y$, $x=1$ and $y=1$.
Prove that the affine equation of the regular conics $\mathcal{D}_{t}$ such that they pass through $Q$ and $R$ and are tangent to $\mathcal{C}$ in $P$ is $xy+t(y-x)=0\quad(t\neq0)$.
Prove that $\mathcal{C}$ and $\mathcal{D}_{t}$ intersect at most in another point. Compute the multiplicity of intersection in each point depending on the values of $t$.
For the first part, I know that the roots of the polynomial are given by $Q$, $R$ and $P$ (double). But I don't know how to obtain the equation.
For the second part, let $F$, $G$ be the following homogeneous polynomials: $$F=Y^2 Z^2 + X^2 Z^2 + X^2 Y ^2 - 2XYZ(X+Y+Z)$$ $$G=XY-tZ(Y-X)$$
I have a problem here, because the usual algorithm we follow to compute intersections need that $(1:0:0)\notin \mathcal C \cap \mathcal D _t$ or $(0:1:0)\notin \mathcal C \cap \mathcal D _t$, so we can use the resultant to find the coordinates. But we have none of these conditions.
For the first part, assume that the conic $D$ is given (in projective coordinates) by the equation $aX^2+bY^2+cZ^2+dXY+eXZ+fYZ=0$. The fact that $D$ passes through $P,Q,R$ implies that $a,b,c = 0$. The fact that $D$ is tangent to $C$ at $P$ implies that the tangent line of $D$ at $P$ is also given by $Y=X$, which implies that $e=-f$.
This almost proves the entire claim, the only thing you need to check is that $d$ cannot be zero, hence you can divide by $d$ to get the equation you want, and also that $e$ cannot be zero. But note that if either $d$ or $e=0$, then the conic you get is not regular.