If a tangent of slope 2 of the ellipse x2a2+y2b2=1 is normal to the circle x2+y2+4x+1=0, then the maximum value of ab is

Equation of tangent of x2a2+y2b2=1 with slope m is y=mx±a2m2+b2⇒y=2x±4a2+b2 It is normal to the circle x2+y2+4x+1=0 hence, this tangent  passes through centre (−2,0)⇒0=−4±4a2+b2⇒4a2+b2=16 Using Arithmetic mean (A.M.) ≥ Geometric mean (G.M.)  we get 4a2+b22≥4a2b2⇒ab≤4  Minimum value of  ab is 4  Therefore, the correct answer is 0004.00