Tangency and Parameter Conditions

1. Set the equations equal and rearrange.
   $$x^2-4x+3=mx-6$$
   $$x^2-(4+m)x+9=0$$
2. Tangent means $\Delta=0$.
   $$(4+m)^2-36=0$$
   $$4+m=\pm6$$
   Therefore $m=2$ or $m=-10$.
3. Substitute $m=2$ and $m=-10$ back to find the contact points.
   $$m=2:\quad (x-3)^2=0$$
   $$x=3,\quad y=0$$
   $$m=-10:\quad (x+3)^2=0$$
   $$x=-3,\quad y=24$$

Show that $y=3x-2k$ and $y=x^2-kx+2$ meet for every value of $k$ (9709/12/F/M/23 Q1).

1. $$\begin{aligned}x^2-(k+3)x\\ {}+(2+2k)&=0\end{aligned}$$
2. $$\Delta=(k+3)^2-8(1+k)$$
   $$\Delta=k^2-2k+1$$
   $$\Delta=(k-1)^2$$

Since $(k-1)^2\ge0$, there are real roots for every $k$; therefore the line and curve always meet.

1. $$2x^2+5=mx-3$$
   $$2x^2-mx+8=0$$
2. Do not meet means $\Delta<0$.
   $$(-m)^2-4(2)(8)<0$$
   $$m^2-64<0$$

For $2x^2-kxy+2=0$ and $y=4x+3$, find the values of $k$ for which the line and curve do not meet. Substitution gives (9709/11/M/J/25 Q6(b)):

The combined equation would become linear at $k=\tfrac12$, because its $x^2$ coefficient would be zero. This value lies outside the final interval, so no separate linear check is needed here.