A) an ellipse whose eccentricity is \[\frac{1}{\sqrt{r}+1}\] where \[r>1\]
B) a hyperbola whose eccentricity is \[\frac{2}{\sqrt{r+1}}\], when \[0<r<1\]
C) a hyperbola whose eccentricity is \[\frac{2}{\sqrt{1-r}}\]when \[0<r<1\]
D) an ellipse whose eccentricity is \[\sqrt{\frac{2}{r+1}}\] when \[r>1\]
Correct Answer: D
Solution :
\[\frac{{{y}^{2}}}{1+r}-\frac{{{x}^{2}}}{1-r}\,=\,1\] If r \[\in \] (0, 1) then this curve is a hyperbola whose \[e=\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}}\,=\,\sqrt{1+\frac{1-r}{1+r}}=\,\,\sqrt{\frac{2}{1+r}}\] If \[r>1\] then this curve is an ellipse whose \[e=\sqrt{1-\frac{{{b}^{2}}}{{{a}^{2}}}}\,=\,\sqrt{1-\left( \frac{r-1}{r+1} \right)}\,=\,\sqrt{\frac{2}{r+1}}\,\,\]
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