How to get the efficiency of a heat engine which undergoes an elliptical cycle?
Let $(V_0,P_0)$ represent the coordinates of the center of the ellipse, and let $P_\mathrm{max}$ and $V_\mathrm{max}$ represent the maximum pressure and maximum volume, respectively over the cycle.  Then, the work is the area of the ellipse, which is given by:  $$W=\pi(P_\mathrm{max}-P_0)(V_\mathrm{max}-V_0)$$This is, $\pi$ times the product of the semi-major and semi-minor axes.  This differs from the result which you gave.

The shape of the ellipse can be represented parametrically in terms of the angle $\theta$ around the cycle, assuming $\theta$ is measured clockwise from the point $(V_0-(V_0-V_\mathrm{max}),P_0)$:
$$P=P_0+(P_\mathrm{max}-P_0)\sin{\theta}\tag{1a}$$
$$V=V_0-(V_\mathrm{max}-V_0)\cos{\theta}\tag{1b}$$
Application of the first law of thermodynamics to the working fluid over a differential portion of the cycle gives us:
$$dU=nC_vdT=dQ-PdV\tag{2}$$But from the ideal gas law, $$nRdT=d(PV)$$Substitution of this into Eqn. 2 yields:
$$\frac{C_v}{R}d(PV)=dQ-PdV$$
So the differential heat added during an arbitrary portion of the cycle is given by:$$dQ=\frac{1}{R}[(C_v+R)PdV+C_VVdP]\tag{3}$$So the differential heat is zero when dQ = 0, or, equivalently, when $d\ln{P}+ d\ln{V^{\gamma}}=0$, or, equivalently when $PV^{\gamma}=\mathrm{Const}$.  This is the equation for an isentropic line tangent to the ellipse.  

There are two angles $\theta$ at which adiabats are tangent to the ellipse.  These two angles can be obtained by substituting Eqns. 1 into Eqn. 3 with dQ = 0.  Once these angles have been determined, the total positive heat added Q during the cycle can then be obtained by integrating Eqn. 3 with respect to $\theta$ between the two angles.