A) \[|a|\le \frac{1}{\sqrt{2}}\]
B) \[a<-\frac{1}{\sqrt{2}}\]
C) \[a>-\frac{1}{\sqrt{2}}\]
D) none of these
Correct Answer: A
Solution :
[a] : \[y=4{{x}^{2}}\Rightarrow {{x}^{2}}=\frac{1}{4}y\] \[\therefore \]\[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{16}=1\]becomes\[\frac{y}{4{{a}^{2}}}-\frac{{{y}^{2}}}{16}=1\] \[\Rightarrow \]\[4y-{{a}^{2}}{{y}^{2}}=16{{a}^{2}}\Rightarrow {{a}^{2}}{{y}^{2}}-4y+16{{a}^{2}}=0\] \[\Rightarrow \]\[D\ge 0\]for intersection of two curves \[\Rightarrow \]\[16-4{{a}^{2}}(16{{a}^{2}})\ge 0\Rightarrow 1-4{{a}^{4}}\ge 0\] \[\Rightarrow \]\[{{(2{{a}^{2}})}^{2}}\le 1\Rightarrow |\sqrt{2}a|\le 1\Rightarrow -\frac{1}{\sqrt{2}}\le a\le \frac{1}{\sqrt{1}}\]
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