Let S = {(x,y) in mathbb{R}^2 : frac{y^2}{1+r | vedclass

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(B) The given equation is $\frac{y^2}{1+r} - \frac{x^2}{1-r} = 1$.Case $1$: If $r > 1$,then $1-r  0$. The equatio

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an ellipse whose eccentricity is $\sqrt{\frac{2}{r+1}}$,when $r > 1$.

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(B) The given equation is $\frac{y^2}{1+r} - \frac{x^2}{1-r} = 1$.Case $1$: If $r > 1$,then $1-r  0$. The equation becomes $\frac{y^2}{1+r} + \frac{x^2}{r-1} = 1$,which is an ellipse.The eccentricity $e$ of an ellipse $\frac{y^2}{a^2} + \frac{x^2}{b^2} = 1$ (where $a^2 > b^2$) is $e = \sqrt{1 - \frac{b^2}{a^2}}$.Here $a^2 = 1+r$ and $b^2 = r-1$. Thus,$e = \sqrt{1 - \frac{r-1}{r+1}} = \sqrt{\frac{r+1-r+1}{r+1}} = \sqrt{\frac{2}{r+1}}$.Case $2$: If $0  0$. The equation is of the form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$,which is a hyperbola.The eccentricity $e$ of a hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ is $e = \sqrt{1 + \frac{b^2}{a^2}}$.Here $a^2 = 1+r$ and $b^2 = 1-r$. Thus,$e = \sqrt{1 + \frac{1-r}{1+r}} = \sqrt{\frac{1+r+1-r}{1+r}} = \sqrt{\frac{2}{1+r}}$.Comparing with the options,option $B$ is correct.

Let $S = \{(x,y) \in \mathbb{R}^2 : \frac{y^2}{1+r} - \frac{x^2}{1-r} = 1\}$,where $r \neq \pm 1$. Then $S$ represents

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