tangent point between 2 parabolas (parameter)

Question

I have 2 parabolas: $P1: y=x^{2}+5x+4 $ and $P2: y=(m-1)x^{2}+(4m+n-4)x+5m+2n-4$ with $m,n$ real numbers and $m$ different from 1.

I need to find parameters $m,n$ such that these 2 parabolas are tangent in $T(-2,-2)$The right answer is $m=-2;n=1$

I started like this: $P1=P2$ and I got a quadratic equation $(2-m)x^{2}+(9-4m-n)x+8-5m-2n=0$

I put the condition that the discriminant to be 0 but the calculation are too heavy.How to continue?

Answer 1

From the fact that $(-2,-2)$ is on $P2$,

$-2=(m-1)4+(4m+n-4)(-2)+5m+2n-4=m.$

Answer 2

Solve the following system: $$P_2(-2)=-2$$ and $$P_2'(-2)=P_1'(-2).$$