‌y2=4x ‌(x−4)2+y2=16 Let equation of tangent of parabola y=mx+1∕m Now equation 1 also touches the circle ‌∴|‌
4m+1∕m
√1+m2
|=4 ‌(4m+1∕m)2=16+16m2 ‌16m4+8m2+1=16m2+16m4 ‌8m2=1 ‌m2=1∕8‌‌{m4=0}(m→∞) For distinet points consider only m2=1∕8. Point of contact of parabola ‌P(8,4√2) ‌∴PQ=√S1⇒(PQ)2=S1 ‌=16+32−16=32