Find the two points on the curve $y=x^4-2x^2-x$ that have a common tangent line.
My solution: Suppose that these two point are $(p,f(p))$ and $(q,f(q))$ providing that $p \neq q$. Since they have a common tangent line then: $y'(p)=y'(q),$ i.e. $4p^3-4p-1=4q^3-4q-1$ and after cancellation we get: $p^2+pq+q^2=1$.
Tangent lines to curve at points $(p,f(p))$ and $(q,f(q))$ are $y=y(p)+y'(p)(x-p)$ and $y=y(q)+y'(q)(x-q)$, respectively. I have tried to put $x=q$ in the first and $x=p$ in the second equations but my efforts were unsuccesfull.
Can anyone explain me how to tackle that problem?
Best Answer
Tangent: $ f(p) + f'(p)(x-p) = f(q) + f'(q)(x-q) $
It follows:
$(1)\,$ : $\enspace f'(p) = f'(q)\enspace => \enspace p^3-q^3 = p-q $
$(2)\,$ : $\enspace f(p)-pf'(p) = f(q)-qf'(q)\enspace => \enspace 3(p^4-q^4) = 2(p^2-q^2) $
$(1)$ into $(2)$ leads to $\enspace 3(p^4-q^4) = 2(p+q)(p-q) = 2(p+q)(p^3-q^3)$
which is equivalent to $\enspace (p-q)^2 (p^2-q^2) = 0\,$ .
Because of the condition $\,p\neq q\,$ we get $\,q=-p\,$ .
Putting this result into $(1)$ and choosing $p>0$ we get $p=1$ and therefore $q=-1$ .
The points are $\,(1;f(1))=(1;-2)\,$ and $\,(-1;f(-1))=(-1;0)\,$ .