Let $C:\frac{y^2}3-x^2=1$, and $F~(0,-2)$. Let $F\in l$, and $l\cap C=A$, $B$. Let $P~\left(0,\sqrt7\right)$. If the center of $(ABP)$ is on the $x$-axis. Find the slope of $l$.
I can let $l:y=kx-2$ because obviously $l:x=0$ doesn't work. Let $A~(x_1,kx_1-2)$, $B~(x_2,kx_2-2)$. Putting the functions of $l$ and $c$ together, we get \[(k-3)x^2-4kx+1=0.\tag1\] And $x_{1,2}$ are the roots of this quadratic function. In addition let $O~(t,0)$ the center of $(ABP)$. So $AO=BO=PO$. By Pythagorean theorem, \[(x_1-t)^2+(kx_1-2)^2=(x_2-t)^2+(kx_2-2)^2=t^2+7.\] Simplify to get that $x_{1,2}$ are the roots of the quadratic function \[(k^2+1)x^2-(4k+2t)x-3=0.\tag2\] Since $(1)$ and $(2)$ have the same roots, Vieta theorem yields \[x_1x_2=\frac1{k-3}=\frac{-3}{k^2+1}.\] But solving this doesn't give me the correct answer. There must be some mistake in my solution. Where?
There is an error in the following step :
It should be $$(\color{red}{k^2}-3)x^2-4kx+1=0$$
Then, you'll finally get $$\frac{1}{k^2-3}=\frac{-3}{k^2+1}$$ Solving this gives $k=\pm\sqrt 2$.