For coax to work at this close a frequency (~7% frequency difference) you need a pretty high-Q trap, which means very low-loss coax. Offhand, having done this before, I don't think "small" coax will work well; 7/8" Heliax or so will work better. 1-5/8" Heliax would work even better than that.
You can make the trap 1/4-wavelength with the end "open," or 1/2-wavelength with the end "shorted." Since 1/4-WL is shorter and will have less loss I'd try the open-end stub first.
Formula would be 246/f(MHz) * VF = stub length in feet. For 134.675 MHz, and assuming the use of very low loss cellular dielectric cable with a VF of 0.84, this would be 246/134.675 * 0.84 = 1.53436 feet, or 18.4123 inches (46.76 cm). The stub length must include the length of the connector used to place it in parallel with the antenna line via a coaxial "tee" adapter. The "open end" should be as close to perfectly open as possible while maintaining coaxial design; I just use a hacksaw or equivalent on the Heliax to cut through everything (jacket, solid copper outer conductor, foam dielectric, copper center conductor) all in one cut to make the end "square," and then clean the end of any materials (shavings). Once the "magic" dimension has been found for maximum notching at the desired frequency, I put a piece of heat shrink tubing over the end and shrink it.
A good high-Q stub can likely work; in my experience, a lower-Q stub, like one made of small flexible coax (.195" or .240" stuff) usually doesn't because the notch isn't deep enough.
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