$a\le c\le 5a$

Hint

The parabola $y = ax^2 + bx + c$ intersects a line $y = mx + d$ in at most one point if the discriminant of the quadratic equation obtained by equating the two is zero or negative. Substituting $y = mx + d$ into $y = ax^2 + bx + c$, we get: \[ ax^2 + (b - m)x + (c - d) = 0. \]The discriminant of this quadratic equation is: \[ \Delta = (b - m)^2 - 4a(c - d). \]For the line to intersect the parabola in at most one point, we require $\Delta \leq 0$, i.e., \[ (b - m)^2 \leq 4a(c - d). \]We analyze each of the six lines using this tangency condition: 1. Line $y = ax + b$: Substituting $m = a$ and $d = b$ into the discriminant condition: \[ (b - a)^2 \leq 4a(c - b). \] 2. Line $y = bx + c$: Substituting $m = b$ and $d = c$: \[ (b - b)^2 \leq 4a(c - c) \implies 0 \leq 0. \]This is always satisfied and imposes no restriction. 3. Line $y = cx + a$: Substituting $m = c$ and $d = a$: \[ (b - c)^2 \leq 4a(c - a). \] 4. Line $y = bx + a$: Substituting $m = b$ and $d = a$: \[ (b - b)^2 \leq 4a(c - a) \implies 0 \leq 4a(c - a). \]Thus, $c \geq a$ if $a > 0$, or $c \leq a$ if $a < 0$. 5. Line $y = cx + b$: Substituting $m = c$ and $d = b$: \[ (b - c)^2 \leq 4a(c - b). \] 6. Line $y = ax + c$: Substituting $m = a$ and $d = c$: \[ (b - a)^2 \leq 4a(c - c) \implies (b - a)^2 \leq 0 \implies b = a. \] From condition 6, we know that $b = a$. Substituting $b = a$ into the other conditions, we get: 1. From Condition 1: \[ (a - a)^2 \leq 4a(c - a) \implies 0 \leq 4a(c - a). \]This implies $c \geq a$ if $a > 0$, or $c \leq a$ if $a < 0$. 2. From Condition 3: \[ (a - c)^2 \leq 4a(c - a). \]Expanding $(a - c)^2 \leq 4a(c - a)$, we get: \[ a^2 - 2ac + c^2 \leq 4ac - 4a^2. \]Simplifying: \[ c^2 - 6ac + 5a^2 \leq 0. \]Factoring: \[ (c - 5a)(c - a) \leq 0. \] The inequality $(c - 5a)(c - a) \leq 0$ holds when $c$ lies between $a$ and $5a$, i.e., \[ a \leq c \leq 5a. \] Dividing through by $a$ (and noting that $a \neq 0$), we obtain: \[ 1 \leq \frac{c}{a} \leq 5. \] Thus, the maximum value of $\frac{c}{a}$ is $\boxed{5}$, and the minimum value is $\boxed{1}$.