BITSAT 2005 Solved Question Paper

Given curve is y2=4x .........(i)
Let the equation of line by y=mx+c
Since

dy

dx

=m=1 and this line is passing through the point (0,1).
∴1=1(0)+c⇒c=1
∴y=x+1 .......(ii)
Solving Eqs. (i) and (ii), we get
(x+1)2=4x
⇒(x−1)2=0
⇒x=1 and y=2
This shows that line touch the curve at one point. So length of intercept is zero.