Let S = \ (x, y) ∈ R^ 2 : (y2)/(1 + r) - (x2)/(1 - r) = 1\ , where r ≠ ± 1 . Then S represents:

Correct answer: C. An ellipse whose eccentricity is (2)/(r + 1) , when r > 1 Given y^ 2 1 + r - x^ 2 1 - r = 1 If r ∈ (0, 1) then this curve is a hyperbola whose eccentricity is 1 + (1 - r)/(1 + r) = (2)/(r + 1) . If r > 1 , then this curve is an ellipse whose eccentricity is 1 - (r - 1)/(r + 1) = (2)/(r + 1) .

JEE Main

Exam: JEE Main
Subject: Mathematics
Chapter: Ellipse

JEE Main · Mathematics · Ellipse

Let S = \ (x, y) ∈ R^ 2 : (y2)/(1 + r) - (x2)/(1 - r) = 1\ , where...

Let

S = \left\{\left(x, y\right) \in R^{2}: \dfrac{y2}{1 + r} - \dfrac{x2}{1 - r} = 1\right\},

where

r \ne \pm 1 .

Then

S

represents:

Options

AAn ellipse whose eccentricity is $\dfrac{1}{\sqrt{r + 1}},$ when $r > 1 .$
BA hyperbola whose eccentricity is $\dfrac{2}{\sqrt{r + 1}},$ when $0 < r < 1 .$
CAn ellipse whose eccentricity is $\sqrt{\dfrac{2}{r + 1}},$ when $r > 1$
DA hyperbola whose eccentricity is $\dfrac{2}{\sqrt{1 - r}},$ when $0 < r < 1$

Correct answer

C. An ellipse whose eccentricity is $\sqrt{\dfrac{2}{r + 1}},$ when $r > 1$

Step-by-step solution

Given

\dfrac{y^{2}}{1 + r} - \dfrac{x^{2}}{1 - r} = 1

r \in

\left(0, 1\right)

then this curve is a hyperbola whose eccentricity is \sqrt{1 + \dfrac{1 - r}{1 + r}} = \sqrt{\dfrac{2}{r + 1}}.<br/><br/>If

r > 1

, then this curve is an ellipse whose<br/>eccentricity is

\sqrt{1 - \dfrac{r - 1}{r + 1}} = \sqrt{\dfrac{2}{r + 1}}.

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