2017-I-18

Therefore, we have

$\begin{array}{rcl}
\Delta & = & (-2k+2)^2 – 4(2)(-3k-11) \\
& = & 4k^2 – 8k + 4 +24k + 88 \\
& = & 4k^2 + 16k + 92 \\
& = & 4(k^2 + 4k + 23) \\
& = & 4(k^2 + 4k + 4 + 19) \\
& = & 4[(k + 4)^2 +19] \\
& = & 4(k+4)^2 + 76
\end{array}$

For all real constant $k$,

$\begin{array}{rcl}
(k+2)^2 & \ge & 0 \\
4(k+2)^2 + 76 & > & 0
\end{array}$

Therefore, $L$ and $\Gamma$ intersect at two distinct points.

1. Note that $a$ and $b$ are the two roots of the equation $(*)$, then we have$\begin{array}{rcl}
   a + b & = & -\dfrac{-2k+2}{2} \\
   a + b & = & k-1
   \end{array}$

   and also

   $\begin{array}{rcl}
   ab & = & \dfrac{-3k-11}{2}
   \end{array}$

   Hence, we have

   $\begin{array}{cl}
   & (a-b)^2 \\
   = & a^2 – 2ab + b^2 \\
   = & a^2 +2ab + b^2 -4ab \\
   = & (a+b)^2 – 4ab \\
   = & (k-1)^2 – 4 \times ( \dfrac{-3k-11}{2} ) \\
   = & (k^2 -2k + 1) + 6k + 22 \\
   = & k^2 +4k + 23
   \end{array}$

2. The distance between $A$ and $B$$\begin{array}{cl}
   = & \left| a -b \right| \\
   = & \sqrt{(a-b)^2} \\
   = & \sqrt{k^2 + 4k + 23} \\
   = & \sqrt{k^2 + 4k + 4 + 19} \\
   = & \sqrt{(k+2)^2 + 19} \\
   \ge & \sqrt{19} \text{, for any real constant $k$} \\
   > & 4
   \end{array}$

   Therefore, it is impossible that the distance between $A$ and $B$ less than $4$.