Find the range of values of m for which the straight line $y+mx=12$ intersects the curve $x^2+xy=12$ at two distinct points.
I am able to get that $m<4$, but according to the answer key, it should be $m<4, m\neq 1$. I do not understand why $m=1$ is not valid, can anyone explain?
Thanks!
Because the intersection is requested at two distinc point.
For $m=1$:
$$x^2+xy=12 \implies x^2+x(12-x)=12\implies x=1$$
$$y+mx=12\implies y=12-mx$$ so $$x^2+xy=x^2+x(12-mx)=12\implies (1-m)x^2+12x-12=0$$ Thus $$x=\frac{-12\pm\sqrt{12^2-4(1-m)(-12)}}{2(1-m)}$$ If $m=1$ then $x$ is undefined.
Note quadratic then becomes $12x-12=0\implies x=1$ - not two points.
Transform the equations by putting $z=x+y$ so that $$xz=12$$ and $$z+(m-1)x=12$$ Then $$z^2+12(m-1)=12z$$
If $m=1$ we have $z=0$ or $z=12$. The second is fine, but the first is not a solution - the point of intersection has "gone to infinity" parallel to the asymptote $z=0$ or $x+y=0$
Recommended to draw a diagram in such cases.
