On most general exact solution for Vaidya-Tikekar isentropic superdense star

The energy density of Vaidya-Tikekar isentropic superdense star is found to be decreasing away from the center, only if the parameter K is negative. The most general exact solution for the star is derived for all negative values of K in terms of circular and inverse circular functions. Which can further be expressed in terms of algebraic functions for K = 2-(n/delta)(2) < 0 (n being integer and delta = 1,2,3 4). The energy conditions 0 less than or equal to p less than or equal to alpha rho c(2), (alpha = 1 or 1/3) and adiabatic sound speed condition d rho/dp less than or equal to c, when applied at the center and at the boundary, restricted the parameters K and alpha such that .18 < -K < 2287 and .004 less than or equal to alpha less than or equal to .86. The maximum mass of the star satisfying the strong energy condition (SEC), (alpha = 1/3) is found to be 3.82 M. at K = -2/3, while the same for the weak energy condition (WEC), (alpha = 1) is 4.57 M. at K = ->5/2. In each case the surface density is assumed to be 2 x 10(14) gm cm(-3). The solutions corresponding to K > 0 (in fact K > 1) are also made meaningful by considering the hypersurfaces t= constant as 3-hyperboloid by replacing the parameter R-2 by -R-2 in Vaidya-Tikekar formalism. The solutions for the later case are also expressible in terms of algebraic functions for K = 2-(n/delta)(2) > 1 (n being integer or zero and delta = 1,2,3 4). The cases for which 0 < K < 1 do not possess negative energy density gradient and therefore are incapable of representing any physically plausible star model. In totality the article provides all the physically plausible exact solutions for the Buchdahl static perfect fluid spheres.