)(c,ec)=ec ⇒ Tangent at (c,ec) y−ec=ec(x−c) it intersect x-axis Put y=0⇒x=c−1 ......(1) Now y2=4x⇒
dy
dx
=
2
y
⇒(
dy
dx
)(1,2)=1 ⇒ Slope of normal = –1 Equation of normal y−2=−1(x−1) x+y=3 it intersect x -axis Put quad‌y=0⇒x=3 .....(2) Points are same ⇒x=c−1=3 ⇒c=4