In jfet, why Vgs is half the cutoff voltage when the source resistor is same as the ohmic resistance of the jfet?
JFETs have a square-law current versus gate voltage.

thus: \$ I_d = \kappa \times (V_g - V_t)^2 \$

the derivative \$ \frac{d I}{d V_g} = \kappa \times [2 \times (V_g - V_t) ] \$

Which tells us what?

The ohmic resistance of the FET is the transconductance, 

or \$ \frac{1}{d I / d V_g}\$

thus ohmic resistance = \$ \frac{1}{[\kappa \times 2 \times (V_g - V_t)]} \$

Is that "2" of use in understanding?

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becca what did you learn? does the math work?