Prove there are no 3 collinear points on a non-degenerate conic

Question

Obviously, I know it is not possible, but I don't know how to prove it. I tried choosing 3 points, using that the slopes between them are equal, then trying to somehow show that the general form of conics $Ax^{2}+Cy^{2}+Dx+Ey+F=0$ has no root. That didn't help. Maybe a geometrical proof would be easier.

Answer 1

Change coordinates in the plane, to make the line $L$ joining the three points the $x$-axis (equation $y=0$). Let your conic $C$ now have equation $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0.\tag1$$ The $x$-coordinates of the points of $C$ on the $x$-axis satisfy $$Ax^2+Dx+F=0.\tag2$$ We get this by putting $y=0$ into $(1)$. But $(2)$ can only have two or fewer solutions, unless all its coefficients are zero, that is $A=D=F=0$. Then $(1)$ becomes $$Bxy+Cy^2+Ey=0\tag3$$ which factorises.

Answer 2

Hint:

Any non-degenerate conic can be reduced to one of the forms

$$x^2\pm y^2=1, \\x^2=y.$$

by an affine transform, that preserves collinearity. So it suffices to consider these three cases.

Take any two distinct points on the conic and intersect the conic with the parametric line that joins them, let $P(t)=(1-t)P_0+tP_1$.

Now if you plug $P(t)$ in one of the above equations, you get a quadratic polynomial in $t$ which must be of the form $\lambda t(1-t)$ with $\lambda\ne0$.