The focal chord of the parabola (y−2)2=16(x−1) is a tangent to the circle x2+y2−14x−4y+51=0, then slope of the focal chord can be

Given parabola is (y−2)2=16(x−1) Here vertex =(h,k)=(1,2)4a=16⇒a=4 Focus =(h+a,k)=(1+4,2)=(5,2) The equation of focal chord is y−2=mx−5⇒mx−y+2−5m=0………...(1) Given circle equation is x2+y2−14x−4y+51=0………...(2) Hence centre (7,2) And radius, r=49+4−51=2Since equation (1) is a tangent of circle equation (2) thenr = d2=|7m−2+2−5m|m2+1(∵d= perpendiculardistance from centre to equation (1))⇒2m2+1=|2m| Squaring on both sides ⇒2m2=m2+1⇒m=±1∴ Slope of focal chord is 1