Question

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

• A. A ellipse whose eccentricity is $\sqrt{\frac{2}{r+1}}$, when $r>1$
• B. A ellipse whose eccentricity is $\frac{1}{\sqrt{r+1}}$, when $r>1$
• C. A hyperbola whose eccentricity is $\sqrt{\frac{2}{1-r}}$, when $-1<r<1$
• D. A hyperbola whose eccentricity is $\sqrt{\frac{2}{r+1}}$, when $-1<r<1$

Answer 1

Answer: (A), (D)

A hyperbola whose eccentricity is $\sqrt{\frac{2}{r+1}}$, when $-1<r<1$

Given equation is $\frac{y^2}{1+r}+\frac{x^2}{r-1}=1$
For above equation to represent an ellipse , $r-1>0$ and $r+1>0$
$\Rightarrow r>1$ and $r>-1$
$\Rightarrow r>1$
So, $\forall r>1$ $\frac{y^2}{1+r}+\frac{x^2}{r-1}=1$ represents ellipse with major axis as $y-$ axis.
$\because r+1>r-1$
$\therefore e=\sqrt{1-\left(\frac{r-1}{r+1}\right)}=\sqrt{\frac{2}{r+1}}$
Similarly, for the given equation to represent hyperbola,
$(r-1)(r+1)<0$
$\Rightarrow r\in (-1,1)$
Now, $\forall r\in (-1,1),$ $r-1\in (-2,0)$ and $r+1\in (0,2)$
So hyperbola will be of form $\frac{y^2}{1+r}-\frac{x^2}{1-r}=1$
Now $e=\sqrt{1+\left(\frac{1-r}{1+r}\right)}=\sqrt{\frac{2}{1+r}}$