Given, equation of normal mx−y+c=0 Parabola y2=16x Comparing with general equation of parabola i.e., y2=4ax⇒a=4 Let the coordinates of point P on parabola be (4t2,8t) ∵ Slope of tangent at point P=‌
1
t
Also, slope of tangent =−‌
1
‌ Slope of normal ‌
=−‌
1
m
From Eqs. (i) and (ii), ‌
1
t
=−‌
1
m
or t=−m ∴ Coordinate of point is P(4m2,−8m). Focus of parabola is (a,0) or (4,0). Given, focal distance =40 Using distance formula, √(4m2−4)2+(−8m−0)2=40 or √(m2−1)2+(−2m)2=10
or √(m2−1)2+4m2=10 or √(m2+1)2=10
or m2+1=10 or m2=9 or m=±3 Therefore, coordinate of point P (4×(±3)2,−8(±3))‌‌≡(36,±24) ∵ Point P lies on line mx−y+c=0 then it must satisfy equation of line (±3)(36)−(±48)+c=0 ±(3×36+48)+c=0 or c=±132 or ‌‌|c|=132