Question P.168: a. Find all lines which are tangent to both of the parabolas......

a. There are two such lines: y = 0 and y = 4x − 4.
Suppose the line y = mx + b is tangent to the parabola y = x² at the point P = ($x_1, x^2_1$) and is also tangent to the parabola y = −x² + 4x − 4 at the point Q = ($x_2, −x_2^2 + 4x_2 −4$). Then, from calculus, the slope of the line is
$m = 2x_1 = −2x_2 + 4 .$                   (1)
First suppose that $x_1 ≠ x_2$; then we can also compute the slope from the fact that P and Q are both on the line. This, along with (1), yields

$m={\frac{-x_{2}^{2}+4x_{2}-4-x_{1}^{2}}{x_{2}-x_{1}}}={\frac{-x_{2}^{2}+4x_{2}-4-(2-x_{2})^{2}}{x_{2}-(2-x_{2})}}\\=-{\frac{(x_{2}-2)^{2}}{x_{2}-1}}.$          (2)

From (1) and (2) we find that $x_2 = 0~or~x_2 = 2.~For~x_2 = 0~we~get~x_1 = 2$; the tangent line to y = x² at (2, 4) is y = 4x − 4, which is also tangent to y = −x² + 4x − 4 at (0, −4). For $x_2 = 2$ we get $x_1 = 0$; the tangent line to y = x² at (0, 0) is y = 0, which is also tangent to y = −x² + 4x − 4 at (2, 0).
On the other hand, if $x_1 = x_2$, equation (1) implies that $x_1 = x_2 = 1$, P = (1, 1), and Q = (1, −1). That is to say, the line joining P and Q is the vertical line x = 1, which is not the tangent line at either point, a contradiction. This completes the consideration of all possibilities.

b. If f(x) = g(x), there will obviously be infinitely many common tangent lines to y = f(x) and y = g(x). Otherwise, there will be either zero, one, or two common tangent lines. If the two parabolas open in the same direction (i.e., both upward or both downward), then there will be no common tangent line if the parabolas do not intersect, one common tangent line if they intersect in exactly one point, and two common tangent lines if the two parabolas intersect in two distinct points. On the other hand, if the two parabolas open in opposite directions, then there will be two common tangent lines if the parabolas do not intersect, one if they intersect
in exactly one point, and none if they intersect in two distinct points. (Note: The parabolas intersect in at most two points since f(x) − g(x) is a nonzero polynomial of degree ≤ 2, so f(x)−g(x) = 0 can have at most two solutions.)
To prove these assertions, note that we can shift the coordinate axes so the vertex of y = f(x) is at the origin; also, replacing f(x) by −f(x) and g(x) by −g(x) if necessary, we can assume that the parabola y = f(x) opens upward. Thus, f(x) = αx² for some α > 0. But now, by replacing x by $x/\sqrt{α}$ (a change of scale along the x-axis), we may assume that f(x) = x². (Of course, g(x) will change accordingly, but it will still be a quadratic polynomial, and the number of common tangent lines and the number of intersection points will not change.)

So we can assume f(x) = x², g(x) = Ax² + Bx + C, A ≠ 0, g(x) ≠ f(x). Let P = ($x_1, x^2_1$) be a point on y = f(x) and Q = ($x_2, Ax_2^2 + Bx_2 + C$) be a point on y = g(x). As in (a), if m denotes the slope of a common tangent line, we have
$m = 2x_1 = 2Ax_2 + B,$                (1)
and if $x_1 ≠ x_2$,

$m=\frac{A x_{2}^{2}+B x_{2}+C-x_{1}^{2}}{x_{2}-x_{1}}\\=\frac{(A-A^{2})x_{2}^{2}+(1-A)B x_{2}+(C-B^{2}/4)}{(1-A)x_{2}-B/2}.$                          (2)

Case 1. Suppose that $x_1 = x_2$. In this case P and Q must be the same point, because otherwise the line connecting them would be vertical and could not be the tangent line. So we have

$x_{1}^{2}=A x_{2}^{2}+B x_{2}+C.$                (3)

Substituting $x_1 = x_2$ into equations (1) and (3) yields

$(A-1)x_{2}+{\frac{B}{2}}=0$                        (4)

and

Subtracting $x_2$ times (4) from this last equation yields

${\frac{B}{2}}\,x_{2}+C=0.$                                (5)

Equations (4) and (5) will have a simultaneous solution if B = C = 0 or if B ≠ 0 and (A − 1)C − B²/4 = 0. In the first of these cases we see that $g(x) = Ax^2$, A ≠ 1, and so we find the unique common tangent line y = 0, at (0, 0). In the second case, we have a common tangent line of slope −4C/B at (−2C/B, 4C²/B²).

Case 2. If $x_1 ≠ x_2$, combining (1) and (2) yields

$(A^{2}-A)x_{2}^{2}+A B x_{2}+\left({\frac{B^{2}}{4}}+C\right)=0.$                                          (6)

If A = 1, then B ≠ 0 from (1), since $x_1 ≠ x_2$. We find that there is a unique solution, with $x_2 = −B/4 − C/B~and~x_1 = B/4 − C/B$. If A ≠ 1, then the quadratic equation (6) for $x_2$ will have two, one, or no solutions depending on whether its discriminant ∆ = A(B² − 4C(A − 1)) is positive, zero, or negative. Note that if there is a common tangent line under Case 1 , then either B = C = 0 or B² = 4C(A − 1); in both cases ∆ = 0. If A ≠ 1 and ∆ > 0, there will be exactly two common tangent lines (corresponding to the two solutions for $x_2$), while if ∆ < 0 there will be none. If A ≠ 1 and ∆ = 0, the double root of (6) will be

using (1), this yield

a contradiction. However, in this case the parabolas have exactly one common tangent line, coming from Case 1.
To summarize, if A ≠ 1, the parabolas have two, one, or no common tangent lines depending on whether ∆ = A (B² − 4C(A − 1)) is positive, zero, or negative. If A = 1, the parabolas have one common tangent line when B ≠ 0 and none when B = 0.
On the other hand, the intersections of the parabolas are given by

If A ≠ 1, this has two, one, or no solutions according to whether
D = B² − 4C(A − 1)
is positive, zero, or negative. Since ∆ = AD, we see that if A > 0 (parabolas open the same way) and A ≠ 1, there are two, one, or zero common tangent lines if there are two, one, or zero intersections, while for A < 0 (parabolas open in opposite directions) it is the other way around. Finally, if A = 1, there is a single intersection point when B ≠ 0 and no intersection when B = 0; the parabolas open the same way, and there are one or zero common tangent lines if there are one or zero intersections, respectively.