Some nice conjectures on conic section Conjecture: Let six points $A,B, C, D, E, F$ lie on a conic. Let $M=AD \cap CF$, $N= BE \cap CF$, $P=BE \cap AD$. Let $M_1=PC \cap ND$, $N_1=PF \cap ME$, $P_1=NA \cap MB$ and $M_2=NA \cap PF$, $N_2=MB \cap PC$, $P_2=ND \cap ME$. Then show that:
1, $ABC$ and $M_1N_1P_1$ are perspective, let the perspector is $X$
2, $ABC$ and $M_2N_2P_2$ are perspective, let the perspector is $Y$
3, $M_1N_1P_1$ and $M_2N_2P_2$ are perspective, let the perspector is $Z$
4, $X, Y, Z$ are collinear

 A: I think you meant $MNP$ instead of $ABC$ in the formulation of 1 and 2.
I managed to prove all conjectures. Since you're not familiar with projective geometry I will try to make the proof as elementary as possible. I will use Menelaos, Ceva, Pascal, Brianchon and Desargues theorems.
First we'll prove the following lemma: Given a triangle $ABC$ and a point $P$. Let $D=BP \cap AC$, $E=CP \cap AB$, $X=AP \cap BC$ and $Y=DE \cap BC$. Then $$\frac{\overrightarrow{BX}}{\overrightarrow{XC}}=-\frac{\overrightarrow{BY}}{\overrightarrow{YC}}.$$

Proof of the lemma: By Ceva theorem we have $$\frac{\overrightarrow{BX}}{\overrightarrow{XC}}\cdot \frac{\overrightarrow{CD}}{\overrightarrow{DA}}\cdot \frac{\overrightarrow{AE}}{\overrightarrow{EB}}=1$$ and by Menelaos theorem $$\frac{\overrightarrow{BY}}{\overrightarrow{YC}}\cdot \frac{\overrightarrow{CD}}{\overrightarrow{DA}}\cdot \frac{\overrightarrow{AE}}{\overrightarrow{EB}}=-1$$
Thus $$\frac{\overrightarrow{BX}}{\overrightarrow{XC}}=\frac{\overrightarrow{DA}}{\overrightarrow{CD}}\cdot \frac{\overrightarrow{EB}}{\overrightarrow{AE}}=-\frac{\overrightarrow{BY}}{\overrightarrow{YC}}.$$
Getting back to the problem, define $N_3=EF \cap AD$, $N_4=NN_1 \cap AD$, $M_3=CD \cap EB$, $M_4=MM_1 \cap EB$, $P_3=AB \cap FC$, $P_4=PP_1 \cap FC$.
Pascal theorem in hexagon $ABEFCD$ says that $M_3, N_3, P_3$ are collinear. 
Menelaos theorem in triangle $MNP$ and line $M_3N_3P_3$ says that $$\frac{\overrightarrow{MP_3}}{\overrightarrow{P_3N}}\cdot \frac{\overrightarrow{NM_3}}{\overrightarrow{M_3P}}\cdot \frac{\overrightarrow{PN_3}}{\overrightarrow{N_3M}}=-1.$$
Lemma says that $$\frac{\overrightarrow{MP_3}}{\overrightarrow{P_3N}} = - \frac{\overrightarrow{MP_4}}{\overrightarrow{P_4N}}, \qquad
\frac{\overrightarrow{NM_3}}{\overrightarrow{M_3P}} = - \frac{\overrightarrow{NM_4}}{\overrightarrow{M_4P}}, \qquad \frac{\overrightarrow{PN_3}}{\overrightarrow{N_3M}} = - \frac{\overrightarrow{PN_4}}{\overrightarrow{N_4M}}.$$
Thus $$\frac{\overrightarrow{MP_4}}{\overrightarrow{P_4N}}\cdot \frac{\overrightarrow{NM_4}}{\overrightarrow{M_4P}}\cdot \frac{\overrightarrow{PN_4}}{\overrightarrow{N_4M}}=1$$
so by Ceva theorem lines $MM_4, NN_4, PP_4$ concur. The conjecture 1. is proved.
Diagonals of hexagon $MN_1PM_1NP_1$ concur so there exists a conic $\mathcal K$ tangent to its sides. 
Consider hexagon $MN_2PM_2NP_2$. Conic $\mathcal K$ is tangent to its sides so in view of Brianchon theorem diagonals $MM_2, NN_2, PP_2$ concur. This yields conjecture 2.
Now take a look at hexagon $M_1N_2P_1M_2N_1P_2$. Again, conic $\mathcal K$ is tangent to its sides so by Brianchon theorem it follows that $M_1M_2, N_1N_2, P_1P_2$ concur. Third conjecture proved.
Observe that $X=MM_1 \cap NN_1$, $Y=MM_2\cap NN_2$ and $Z=M_1M_2\cap N_1N_2$ so collinearity of $X,Y,Z$ is equivalent to perspecivity of triangles $MM_1M_2$ and $NN_1N_2$ and thus by Desargues theorem it is equivalent to concurrence of lines $MN, M_1N_1, M_2N_2$. But this is equivalent to perspectivity of triangles $MN_1M_2$ and $NM_1N_2$ and so by Desargues it is equivalent to collinearity of points $MN_1 \cap NM_1 = P_2$, $N_1M_2 \cap M_1N_2 = P$ and $MM_2 \cap NN_2 = Y$. This is true as we already proved conjecture 2. Thus $X,Y,Z$ are collinear.
