Conic Sections Question 80
Question: The line $ y=mx+c $ touches the parabola $ x^{2}=4ay $ , if
[MNR 1973; MP PET 1994, 99]
Options:
A) $ c=-am $
B) $ c=-a/m $
C) $ c=-am^{2} $
D) $ c=a/m^{2} $
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Answer:
Correct Answer: C
Solution:
$ x^{2}=4a(mx+c) $
therefore $ x^{2}-4amx-4ac=0 $
It touches, then $ B^{2}-4AC=0 $
therefore $ 16a^{2}m^{2}+16ac $
therefore $ \Rightarrow $
therefore $ c=-am^{2} $ .
BETA
