# Let

Question:

Let

$\mathrm{S}=\left\{(x, y) \in \mathbf{R}^{2}: \frac{y^{2}}{1+\mathrm{r}}-\frac{x^{2}}{1-\mathrm{r}}=1\right\}$

where $r \neq \pm 1$ Then $S$ represents:

1. (1) a hyperbola whose eccentricity is $\frac{2}{\sqrt{1-r}}$, when $0 2. (2) an ellipse whose eccentricity is$\sqrt{\frac{2}{r+1}}$, when$r>1$3. (3) a hyperbola whose eccentricity is$\frac{2}{\sqrt{\mathrm{r}+1}}$, when$0<\mathrm{r}<1$4. (4) an ellipse whose eccentricity is$\frac{1}{\sqrt{r+1}}$, when$r>1$Correct Option: , 2 Solution: Since,$r \neq \pm 1$, then there are two cases, when$r>1\frac{x^{2}}{r-1}+\frac{y^{2}}{r+1}=1$(Ellipse) Then,$(r-1)=(r+1)\left(1-e^{2}\right) \Rightarrow 1-e^{2}=\frac{(r-1)}{(r+1)}\Rightarrow \quad e^{2}=1-\frac{(r-1)}{(r+1)}=\frac{2}{(r+1)}\Rightarrow \quad e=\sqrt{\frac{2}{(r+1)}}$When$0

$\frac{x^{2}}{1-r}-\frac{y^{2}}{1+r}=-1$ (Hyperbola)

Then,

$(1-r)=(1+r)\left(e^{2}-1\right) \Rightarrow e^{2}=1+\frac{(r-1)}{(r+1)}=\frac{2 r}{(r+1)}$

$\Rightarrow \quad e=\sqrt{\frac{2 r}{r+1}}$