How can a capacitor in parallel reduce current in an AC circuit.?

A capacitor in parallel with a motor will not change the current or speed of the motor.  It will increase the power factor, but the total power will not change.  Increasing the power factor is good for the power company, but it won't reduce your bill.

Motors acts as loads with inductance. The  motor's inductance draws current at a lagging 90° phase, that is the voltage across the inductor  lead the current through the inductor by 90°. The added capacitor draws current that leads the applied voltage by 90°. The capacitor's current leads the inductor's current by 180°, the capacitor current actually cancels out the inductor's current.

It has something to do with the power triangle.

The apparent power is rated VA (volt ampere).  The reactive power is rated VAR (Volt ampere reactive) and real power is rated is watts
./|    the hypotenuse is the apparent power, the bottom line is real power and the vertical
/_|  line is the  reactive power (i am sorry yahoo wont allow me to properly demonstrate the triangle.)

The reactive power from the inductor is positive from what I can remember and the reactive power created by the capacitor is negative.  They subtract each other.  If the reactive power decreases then the apparent power decreases.  When one side of the triangle decreases, the hypotenuse decreases.  The voltage supplied remains the same regardless but the current will decrease when the apparent power decreases.  The apparent power is the direct multiplication of the voltage supplied and the current supplied.  It is called apparent because it is power that the machine appears to draw.  The machine will draw current during an interval of the cycle and then return it on the rest of the cycle.  The net current in the line per cycle is initially 2.4 amperes.  What the capacitor does is take some of the return current from the machine and store it.  When the machine, needs the current it takes the current from the capacitor and the line.  Since less current has to go back and forth on the line with the introduction of capacitor the current decreases 2.2 amperes.
 
0.5 hp is 373 watts.  373/240 = 1.55 amperes  or 1.6 (I will use this number on my analogy)

If I was to explain this in non technical terms, think of the current as the number of cars in the road. If the auto repair shop takes in 20 car repairs in the morning but needs to have 16 cars in the afternoon so they have more space to work, it will have 20 cars come in the morning and 4 cars get sent back in the afternoon.  The net traffic in the road to the auto shop is 20 cars coming in and 4 cars going back which is a total of 24 if summed up.  Let me point out that the current component of the real power is 1.6 amperes so I would say that by the end of the day the auto shop finish 16 car repairs a day.  Now lets have two parking spot or storage for two cars. The storage would be your capacitor in electrical terms.  So instead of sending 4 cars back because of storage (space) problems,  you are only sending 2 back since 2 can be stored in the shop.  Net traffic 20 come in and 2 get sent back in the afternoon which is equal to 22.  The introduction of storage reduces traffic in the road to the auto shop.  Because of storage, the next day, the shop nows take in 18 cars plus 2 from storage which is a total of 20 (they like to have 20 incoming repairs each day) they send 4 cars back total of 22 traffic.  Yes, I know, it's a very poorly managed shop.

Now you might argue that well those 16 cars that are repaired still needs to use the road to get sent back to the owner.  I would say the repaired one uses a different road.  Because in electric terms power that gets consumed dont go back along the electric line but gets output as useful power.  Well theres losses and stuff after that but lets keep it simple right now.