Question

If it is given that S = \left\\{ \left( x,y \right)\in {{R}^{3}}:\dfrac{{{y}^{2}}}{1+r}-\dfrac{{{x}^{2}}}{1-r}=1 \right\\}, where $r\ne \pm 1$. Then what does S represent?
(a) A hyperbola whose eccentricity is $\dfrac{2}{\sqrt{r+1}}$, where 0 < r < 1.
(b) An ellipse whose eccentricity is $\dfrac{1}{r+1}$, where r > 1.
(c) A hyperbola whose eccentricity is $\dfrac{2}{\sqrt{1-r}}$, when 0 < r < 1.
(d) An ellipse whose eccentricity is $\sqrt{\dfrac{2}{r+1}}$, when r > 1.

Answer 1

To solve this problem we will have to find the eccentricity of the given curve S. Generally for the curve $\dfrac{{{y}^{2}}}{{{b}^{2}}}+\dfrac{{{x}^{2}}}{{{a}^{2}}}=1$ eccentricity(e) is given by e = $\sqrt{1-\left( \dfrac{{{a}^{2}}}{{{b}^{2}}} \right)}$ if it is given that b > a. If the obtained eccentricity is between 0 and 1, then it is ellipse, if the eccentricity is greater than 1 then it is hyperbola and if we get eccentricity equal to 1 then it will be a parabola.

Complete step by step answer:
We are given that,
S = \left\\{ \left( x,y \right)\in {{R}^{3}}:\dfrac{{{y}^{2}}}{1+r}-\dfrac{{{x}^{2}}}{1-r}=1 \right\\},
So, to find which type of the curve S represents then for that we need to find the eccentricity of the given curve.
We know that the eccentricity(e) of the curve $\dfrac{{{y}^{2}}}{{{b}^{2}}}+\dfrac{{{x}^{2}}}{{{a}^{2}}}=1$ is given by e = $\sqrt{1-\left( \dfrac{{{a}^{2}}}{{{b}^{2}}} \right)}$ where b > a.
So in our given curve S if we take $1-r$ as $-{{a}^{2}}$ and $1+r$ as ${{b}^{2}}$, then we get the eccentricity of the given curve as,
e = $\sqrt{1-\dfrac{r-1}{r+1}}$
taking LCM of the equation we get,
e = $\sqrt{\dfrac{\left( r+1 \right)-\left( r-1 \right)}{r+1}}$
e = $\sqrt{\dfrac{2}{r+1}}$
Hence now if we observe and restrict the value of r as r > 1 then it will be an ellipse because then e < 1 and we know that when eccentricity is less than one then the curve is an ellipse.
Hence eccentricity is $\sqrt{\dfrac{2}{r+1}}$ and the given curve is an ellipse.
So, option (d) is the correct answer.

Note:
You may make a mistake while finding the eccentricity i.e. we have let $1-r$ as $-{{a}^{2}}$ and not as ${{a}^{2}}$ because in the given curve S we have minus sign in between the equation instead of positive so to compensate that we have taken minus common from $1-r$ and then applied the general formula of eccentricity.