A) \[x=\pm 1\]
B) \[y=\pm 2\]
C) \[y=\pm \sqrt{2}\]
D) \[x=\pm \sqrt{3}\]
Correct Answer: C
Solution :
\[{{(x+1)}^{2}}-{{y}^{2}}-1+5=0\] Þ \[-\frac{{{(x+1)}^{2}}}{4}+\frac{{{y}^{2}}}{4}=1\] Equation of directrices of \[\frac{{{y}^{2}}}{{{b}^{2}}}-\frac{{{x}^{2}}}{{{a}^{2}}}=1\] are \[y=\pm \frac{b}{e}\] Here \[b=2,\ e=\sqrt{1+1}=\sqrt{2}\] Hence \[y=\pm \frac{2}{\sqrt{2}}\] Þ \[y=\pm \sqrt{2}\].
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