Find a line which is tangent to the curve $y=x^4-4x^3$ at 2 points

Question

How can I solve this? Should I set the two points as $a^4-4a^3$ and $b^4-4b^3$?

Answer 1

Let $(t,t^4-4t^3)$ be a point of contact, then

$$\frac{dy}{dx} = \frac{\dot{x}}{\dot{y}} = 4t^3-12t^2$$

The equation of tangent is

$$(y-t^4+4t^3)=4t^2(t-3)(x-t)$$

Let $(s^4-4s^3,s)$ be the self-intersection of its tangent.

\begin{align*} (s^4-4s^3)-(t^4-4t^3) &= 4t^2(t-3)(s-t) \\ (s^4-t^4)-4(s^3-t^3) &= 4t^2(t-3)(s-t) \\ (s^2-t^2)(s^2+t^2)-4(s-t)(s^2+st+t^2) &= 4t^2(t-3)(s-t) \\ (s-t)[(s+t)(s^2+t^2)-4(s^2+st+t^2)] &= 4t^2(t-3)(s-t) \\ (s-t)[(s+t)(s^2+t^2)-4(s^2+st+t^2)-4t^2(t-3)] &= 0 \\ (s-t)[s^3+(t-4)s^2+t(t-4)s+t^2(8-3t)] &= 0 \\ (s-t)^2[s^2+2(t-2)s+t(3t-8)] &= 0 \\ \end{align*}

The double root $s=t$ is due to tangency, thus another intersections are given by $$s=2-t\pm \sqrt{2(2+2t-t^2)}$$

provided $1-\sqrt{3} \le t \le 1+\sqrt{3}$.

The line will be a common tangent if $\Delta=0$, that is $$(s,t)=(1\pm \sqrt{3},1\mp \sqrt{3})$$

Answer 2

Let $f(x):=x^4-4x^3.$

The tangent to the curve in $(a,f(a))$ has equation:

$$y:=g_a(x)=(4a^3-12a^2)(x-a)+a^4-4a^3$$

(Do you see why ?). Using a graphical aid and making a progressive zoom on the region of interest, by trial and error, I saw that the value of $a$ such that there is two contact points is :

$$a=2.73205...$$

I recognized there the decimals of $\sqrt{3}=1.73205...$. Thus I made the assumption that $a=1+\sqrt{3}$.

Then I computed the equation of the tangent for this particular value of $a$, which is $g_{1+\sqrt{3}}(x)=-8x-4$.

Then I checked that the function obtained by taking the difference between the equation of the curve and the equation of the tangent:

$$(x^4-4x^3)-(-8x-4)=(x^2-2x-2)^2$$

has two double roots (in $x=1\pm \sqrt{3}$.)

Answer 3

Suppose a line $y=mx+c$ is tangent to the curve $y=x^4-4x^3$ at two points. Then the difference between these two equations must be a quartic polynomial with two double roots: $$x^4-4x^3-mx-c=(x-a)^2(x-b)^2$$ $$=x^4-2(a+b)x^3+(a^2+4ab+b^2)x^2-2ab(a+b)x+(ab)^2$$ Comparing coefficients, we have $$2(a+b)=4$$ $$a^2+4ab+b^2=(a+b)^2+2ab=0$$ $$m=2ab(a+b)$$ $$c=-(ab)^2$$ The first two relations yield values for $a+b$ and $ab$: $$a+b=\frac42=2$$ $$ab=-\frac{(a+b)^2}2=-\frac{2^2}2=-2$$ The last two relations allow $m$ and $c$ to be evaluated: $$m=2\cdot-2\cdot2=-8$$ $$c=-(-2)^2=-4$$ Hence the line tangent to $y=x^4-4x^3$ at two points is $y=-8x-4$.