Equation of conics 16x2+4y2=1,a2x2−9y2=1 Equation of eccentricity of an ellipse b2=a2(1−e2) 4=16(1−e2) ⇒e2=1−1/4=3/4 e=±43 Focii of an ellipse =(±ae,0) =(±4⋅23,0)=(±23,0) Given, focii of both conics are coincides. ⇒(±23,0)=(±ae,0) [∵ Here (±ae,0) is focii of second conic.] ⇒±ae=±23 ⇒a2e2=12 Equation of eccentricity of second conic (hyperbola) b2=a2(e2−1) ⇒b2=a2e2−a2 ⇒9=12−a2 ⇒a2=3 ⇒a=3