一条直线的方程为 $y=6x-c$，一条曲线的方程为 $y=cx^2+2x-3$，其中 $c$ 为常数。该直线在点 $P$ 处与曲线相切。

求 $c$ 的可能值，以及对应的 $P$ 点坐标。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

相切时联立方程有重根。先排除二次项系数为零的情形，再用判别式确定 $c$。

The intersection equation is $cx^2-4x+c-3=0$. If $c=0$, the graphs are straight lines with different gradients, so they do not touch tangentially. For $c\ne0$, tangency requires

The repeated root is $x=2/c$. When $c=4$, we have $x=1/2$ and $y=6(1/2)-4=-1$. When $c=-1$, we have $x=-2$ and $y=6(-2)+1=-11$. Thus the two pairs are $c=4$ with $P=(1/2,-1)$ and $c=-1$ with $P=(-2,-11)$.

也可以用切点处两图像斜率相同及纵坐标相同，得到两条条件。

At a point of tangency, $2cx+2=6$, so $cx=2$. The intersection condition $cx^2-4x+c-3=0$ then becomes $-2x+c-3=0$. Substituting $x=2/c$ gives $c^2-3c-4=0$, so $c=4$ or $-1$. Substitution in $x=2/c$ and $y=6x-c$ gives $(1/2,-1)$ and $(-2,-11)$ respectively.