If the eccentricity of the hyperbola x2−y2sec2⁡θ=5 is 3 times the eccentricity of the  ellipse x2sec2⁡θ+y2=25, and smallest positive value of θ is πP then twice the value of 'p'is

Eccentricity of the hyperbola x2−y2sec2⁡θ=5 is equation is x25-y25sec2θ=1e1=1+sec2⁡θsec2⁡θ=1+cos2⁡θ Eccentricity of the ellipse x2sec2⁡θ+y2=25 is x225sec2θ+y225=1e2=sec2⁡θ−1sec2⁡θ=|sin⁡θ| Given e1=3e2⇒1+cos2⁡θ=3sin2⁡θ⇒cos⁡θ=±12∴ Least positive value of θ is π4∴P=4⇒2P=8