Tangent Common to two curves, not necessarily at the same point

Question

A common tangent to two curves is a line that is tangent to the two curves, but not necessarily at the same point.

Find, in terms of $a$ and $b$, the explicit equation of the common tangent to the two curves $y = x^2 + ax + b$ and $y = x^2 + bx + a$, where $a$ is not equal to $b$.

Also find, in terms of $a$ and $b$, the $x$ coordinates of the points of contact of this common tangent with each of the two curves.

Answer 1

$$y_1=x_1^2+ax_1+b \tag{1}$$

$$y_2=x_2^2+bx_2+a \tag{2}$$

Equating the slope of the common tangent:

$$m=2x_1+a=2x_2+b$$

$$x_1=\frac{m-a}{2} \tag{3}$$

$$x_2=\frac{m-b}{2} \tag{4}$$

Now \begin{align} m &= \frac{y_1-y_2}{x_1-x_2} \\ &= \frac{x_1^2+ax_1+b-x_2^2-bx_2-a}{x_1-x_2} \\ &= \frac{(m-a)^2+2a(m-a)+4b-(m-b)^2-2b(m-b)-4a}{2(m-a)-2(m-b)} \\ &= \frac{m^2-a^2+4b+b^2-m^2-4a}{2(b-a)} \\ &= \frac{b^2-a^2+4(b-a)}{2(b-a)} \\ &= \frac{a+b}{2}+2 \end{align}

Answer 2

Calling

$$ \left\{ \begin{array}{rcl} f_1(x,y) & = & x^2 + a x + b - y\\ f_2(x,y) & = & x^2 + b x + a - y\\ L& \rightarrow & p = p_1+ \lambda (p_2-p_1) \end{array} \right. $$

with $p = (x,y), \ p_1 = (x_1,y_1) \ p_2 = (x_2,y_2)$ such that $f_1(p_1) = f_2(p_2) = 0$ the tangency problem can be established as:

$$ \left\{ \begin{array}{l} f_1(p_1+\lambda(p_2-p_1)) = 0 \Rightarrow ((a + 2 x_1) (x_1 - x_2) - y_1 + y_2)^2 -4 (x_1 - x_2)^2 (b + x_1 (a + x_1) - y_1)= 0\\ f_2(p_1+\lambda(p_2-p_1)) = 0\Rightarrow ((b + 2 x_1) (x_1 - x_2) - y_1 + y_2)^2 -4 (x_1 - x_2)^2 (a + x_1 (b + x_1) - y_1) = 0 \end{array} \right. $$

because in the corresponding equations due to $f_i = \phi(x_i,y_i)\lambda^2+\psi_i(x_i,y_i)\lambda + \theta_i(x_i,y_i) = 0, \ \ \{i = 1,2\}$ the discriminant associated to the $\lambda$ solution should be null.

Now substituting $y_1 = x_1^2 + a x_1 + b$ and $ y_2 = x_2^2+b x_2 + a$ we obtain

$$ \left\{ \begin{array}{rcl} b - (x_1 - x_2)^2 + a (x_2-1) - b x_2 & = & 0\\ b + a (x_1-1) - b x_1 + (x_1 - x_2)^2 & = & 0 \end{array} \right. $$

and solving for $x_1,x_2$ we get

$$ \left\{ \begin{array}{rcl} x_1 & = & 1/4 (4 - a + b)\\ x_2 & = & 1/4 (4 + a - b) \end{array} \right. $$

and the tangent line reads

$$ L\rightarrow p = \left\{\frac{1}{4} (-a+b+4),\frac{1}{16} \left(-3 a^2+2 a (b+4)+b (b+24)+16\right)\right\}+\lambda \left\{\frac{a-b}{2},\frac{1}{4} (a-b) (a+b+4)\right\} $$ Attached a plot for $a = 2, b = -2$

Answer 3

Before grinding through some generic calculation, it’s a good idea to look for special features of the problem that will simplify it. In this case, the two curves are parabolas that are related by a translation. Any common tangent line must be invariant under this translation, which implies that it must be parallel to the translation direction.

The vertices of the two parabolas are easily found to be at $\left(-\frac a2,b-\frac{a^2}4\right)$ and $\left(-\frac b2,a-\frac{b^2}4\right)$. The difference is $\left(\frac12(b-a),\frac14(b-a)(a+b+4)\right)$ from which we get $\frac12(a+b+4)$ for the slope of the common tangent. From here it’s a fairly straightforward matter of finding where on the curves the derivative has this value and generating the equation of the line from a point and slope or from the two points.