A) \[c=-am\]
B) \[c=-a/m\]
C) \[c=-a{{m}^{2}}\]
D) \[c=a/{{m}^{2}}\]
Correct Answer: C
Solution :
\[{{x}^{2}}=4a(mx+c)\]Þ \[{{x}^{2}}-4amx-4ac=0\] It touches, then \[{{B}^{2}}-4AC=0\] Þ \[16{{a}^{2}}{{m}^{2}}+16ac\]Þ\[\Rightarrow \]Þ\[c=-a{{m}^{2}}\].
You need to login to perform this action.
You will be redirected in
3 sec
