Hyperbola 1 Question 5
6. Let $S=(x, y) \in R^{2}: \frac{y^{2}}{1+r}-\frac{x^{2}}{1-r}=1$,
where $r \neq \pm 1$. Then, $S$ represents (2019 Main, 10 Jan II)
(a) a hyperbola whose eccentricity is $\frac{2}{\sqrt{1-r}}$, when $0<r<1$.
(b) a hyperbola whose eccentricity is $\frac{2}{\sqrt{r+1}}$, when $0<r<1$. (c) an ellipse whose eccentricity is $\sqrt{\frac{2}{r+1}}$, when $r>1$.
(d) an ellipse whose eccentricity is $\frac{1}{\sqrt{r+1}}$, when $r>1$.
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Answer:
Correct Answer: 6. (c)
Solution:
- Given, $S=(x, y) \in R^{2}$ : $\frac{y^{2}}{1+r}-\frac{x^{2}}{1-r}=1$
$$ =(x, y) \in R^{2}: \frac{y^{2}}{1+r}+\frac{x^{2}}{r-1}=1 $$
For $r>1, \frac{y^{2}}{1+r}+\frac{x^{2}}{r-1}=1$, represents a vertical ellipse.
$$ [\because \text { for } r>1, r-1<r+1 \text { and } r-1>0] $$
Now, eccentricity $(e)=\sqrt{1-\frac{r-1}{r+1}}$
$$ \begin{aligned} \because & \text { For } \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a<b, e=\sqrt{1-\frac{a^{2}}{b^{2}}} \\ & =\sqrt{\frac{(r+1)-(r-1)}{r+1}} \\ & =\sqrt{\frac{2}{r+1}} \end{aligned} $$
