$\text{ The coordinates of the point on the parabola }{y}^2=8x\text{ which is at minimum distance from the circle }{x}^2+(y+6{)}^2=1\text{ are }$

detailed solution

Correct option is A

Let   P​(2t2 ,  4t) be any point on the parabola.  The centre of  the given circle is O (0,  -6) and the radius is 1OP2=4t4+(4t+6)2=4t4+4t2+9+12t=4x, where x=t4+4t2+12t+9dxdt=4t3+8t+12=4t3+2t+3=4(t+1)t2−t+3 So dxdt=0⇒t=−1 (other roots are imaginary)  So d2xdt2=43t2+2,d2xdt2t=−1>0 Hence OP2 is minimum at t=−1. But if A is any point on  the circle and on OP (min), then AP will be minimum when  OP is minimum as AP=OP - (radius of circle), Thus the  required point is P(2(−1))2,4(−1)=(2,−4)