A) \[3y-3=\sqrt{6}\]
B) \[3y+3=\sqrt{6}\]
C) \[y+1=\sqrt{3}\]
D) \[y-1=-\sqrt{3}\]
Correct Answer: B
Solution :
Given that, \[8({{x}^{2}}-2x+1)+6({{y}^{2}}+2y+1)\] \[+13-8-6=0\] \[\Rightarrow \] \[\frac{{{(x-1)}^{2}}}{1/8}+\frac{{{(y+1)}^{2}}}{1/6}=1\] Here, \[b>a\] Now, \[e=\sqrt{1-\frac{1/8}{1/6}}=\sqrt{\frac{2}{8}}=\frac{1}{2}\] \[\therefore \] Directrix, \[y+1=\pm \left( \frac{\sqrt{1/6}}{1/2} \right)\,\,\,\left[ \because \,\,y=\pm \frac{b}{e} \right]\] \[\Rightarrow \] \[y+1=\pm \frac{2}{\sqrt{6}}\] \[\Rightarrow \] \[3y+3=\pm \sqrt{6}\]
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