# Line $y=mx+c$ will touch the parabola $x^2=4ay$ if

$\begin{array}{1 1}(A)\;c=am^2 \\(B)\;c=-am^2 \\(C)\;c=2am \\(D)\;none\;of\;these \end{array}$

$x^2=4ay$
$y=mx+c$
From (1) and (2) we get
$x^2=4a(mx+c)$
=> $x^2-4amx-4ac=0$
$D=16a^2m^2+16ac$
$\qquad= 16a(am^2+c)$
Line 2 will touch the parabola if $D=0$
i.e if $C= -am^2$
Hence B is the correct answer.