Is back emf inductance in a ignition coil circuit symmetric for on/off cases?
No, it isn't symmetric in that sense.

In a circuit that consists primarily of inductance and resistance, there's a time constant associated with its transient response1 that works out to

$$\tau = \frac{L}{R}$$

In the igntiion circuit, you basically have a battery, a switch and the ignition coil in series. \$R\$ represents the total resistance of all three. Two of them have constant resistance, but the switch varies between very low and very high resistance.

When the switch is closed, the total \$R\$ is low (dominated by the resistance of the battery and the coil), which means that \$\tau\$ is relatively large &mdash; the current changes more slowly. When the switch is open, the total \$R\$ is extremely large (dominated by the switch), so \$\tau\$ is very small and the current changes very rapidly.

Therefore, since the voltage is determined by 

$$V(t) = L\frac{dI(t)}{dt}$$

when \$\frac{dI}{dt}\$ is large, so is the voltage.



1 The transient response is of the form

$$I(t) = I_{\infty}+(I_0 -I_{\infty})e^{-\frac{t}{\tau}}$$

At t = 0, I(t) = I0 and at t = &infin;, I(t) = I&infin;.