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

We are given the set

$$S=\Bigl\{(x,y)\in\mathbb R^{2}:\;\frac{y^{2}}{1+r}-\frac{x^{2}}{1-r}=1\Bigr\},\qquad r\ne\pm1.$$

First we inspect the signs of the denominators because the nature of the conic depends on them.

For $$r>1$$ we have $$1+r>0$$ and $$1-r<0.$$ To avoid the minus sign in the second fraction we rewrite it:

$$\frac{y^{2}}{1+r}-\frac{x^{2}}{1-r}=1 \;\Longrightarrow\; \frac{y^{2}}{1+r}+\frac{x^{2}}{r-1}=1,$$ because $$\frac{-x^{2}}{1-r}=\frac{x^{2}}{r-1}.$$

Now both denominators are positive, so the equation is of the form

$$\frac{y^{2}}{b^{2}}+\frac{x^{2}}{a^{2}}=1,$$

which is the standard equation of an ellipse. By simple comparison we have

$$a^{2}=r-1,\qquad b^{2}=1+r.$$ (Notice that because $$r>1,$$ we indeed have $$b^{2}>a^{2},$$ so the larger denominator has been kept with the $$y^{2}$$ term.)

For an ellipse, the eccentricity $$e$$ is defined by the relation

$$e=\frac{c}{b},\qquad\text{where }c^{2}=b^{2}-a^{2}.$$

We therefore calculate

$$c^{2}=b^{2}-a^{2}=(1+r)-(r-1)=2.$$

So

$$c=\sqrt{2}.$$

Substituting into the eccentricity formula we get

$$e=\frac{c}{b}=\frac{\sqrt{2}}{\sqrt{1+r}} =\sqrt{\frac{2}{1+r}}.$$

Thus, when $$r>1,$$ the curve is an ellipse whose eccentricity is $$\sqrt{\dfrac{2}{\,r+1\,}}.$$

This matches exactly the description in Option C.

For completeness, note that when $$0<r<1$$ both $$1+r$$ and $$1-r$$ are positive, so the original equation stays in the form

$$\frac{y^{2}}{1+r}-\frac{x^{2}}{1-r}=1,$$

which is the standard vertical hyperbola $$\dfrac{y^{2}}{a^{2}}-\dfrac{x^{2}}{b^{2}}=1.$$ One may show its eccentricity is $$\sqrt{\dfrac{2}{1+r}},$$ which is different from the values quoted in the other options, so none of the hyperbola statements is correct.

Hence, the correct answer is Option C.