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  • Let <img alt="\mathrm{S=\left\{(x, y) \in R^2: \frac{y^2}{1+r}-\frac{x^2}{1-r}=1\right\}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BS%3D%5Cleft%5C%7B%28x%2C%20y%29%20%5Cin%2

Let \mathrm{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
 

Option: 1


\text{a hyperbola whose eccentricity is}\mathrm{\frac{\sqrt{2}}{\sqrt{r+1}}, when 0<r<1.}


Option: 2

\text{ a hyperbola whose eccentricity is} \mathrm{\frac{2}{\sqrt{1-r}}, when 0<r<1.}


Option: 3

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


Option: 4

\text{ an ellipse whose eccentricity is} \mathrm{\sqrt{\frac{2}{r-1}}, when r>1.}


Answers (1)

The given equation is, \mathrm{\frac{y^2}{1+r}-\frac{x^2}{1-r}=1,} where \mathrm{ r \neq \pm 1.}

 \mathrm{ For~ r>1, \frac{y^2}{1+r}+\frac{x^2}{r-1}=1,}

which is equation of an ellipse.

Now, eccentricity, \mathrm{ e=\sqrt{1-\left(\frac{r-1}{r+1}\right)}=\sqrt{\frac{2}{r+1}}}

For \mathrm{ 0<r<1, \frac{y^2}{1+r}-\frac{x^2}{1-r}=1,}

which is equation of a hyperbola.

\mathrm{ \therefore Eccentricity, e=\sqrt{1+\frac{1-r}{1+r}}=\sqrt{\frac{2}{r+1}}}

Posted by

Ramraj Saini

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