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

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

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1 Answer

From (1) and (2) we get
=> $ x^2-4amx-4ac=0$
$\qquad= 16a(am^2+c)$
Line 2 will touch the parabola if $D=0$
i.e if $C= -am^2$
Hence B is the correct answer.
answered Apr 9, 2014 by meena.p

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