1B. The set of parabolas is defined by the formula $y=(k+1) x^{2}-2 k x+k-1, k \in \mathbb{R} \backslash\{-1\}$.

a) Determine the geometric locus of the vertices of these parabolas;

b) Solve the equation $(k+1) x^{2}-2 k x+k-1=0$.
Solution. a) Let $T(x, y)$ be the vertex of the parabola $y=(k+1) x^{2}-2 k x+k-1$, then

$$
x=\frac{k}{k+1}=1-\frac{1}{k+1}, \quad y=-\frac{1}{k+1}
$$

By eliminating $k$ from the above equations, we obtain the desired geometric locus: $y=x-1$. Therefore, the geometric locus of the vertices of the parabolas is a straight line.

b) Since $D=4 k^{2}-4\left(k^{2}-1\right)=4$, we get $x_{1}=1, x_{2}=\frac{k-1}{k+1}$.