Eccentricity of the conic represented by x-ea-e-a2-3- y-ea-e-a2 is

x= e ^ a + e ^ a 2√( 3 ) #160; ,y= e ^ a e ^ a 2 x ^ 2 = e ^ 2a + e ^ 2a +2 12 , y ^ 2 = e ^ 2a + e ^ 2a 2 4 x ^ 2 + y ^ 2 = e ^ 2 ...

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Standard XII

Mathematics

Functions

Eccentricity ...

Question

Eccentricity of the conic represented by $x = \frac{e^{a} + e^{- a}}{2 \sqrt{3}}, y = \frac{e^{a} - e^{- a}}{2}$ is

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Solution

$x = \frac{e^{a} + e^{- a}}{2 \sqrt{3}}, y = \frac{e^{a} - e^{- a}}{2}$
$x^{2} = \frac{e^{2 a} + e^{- 2 a} + 2}{12}, y^{2} = \frac{e^{2 a} + e^{- 2 a} - 2}{4}$
$x^{2} + y^{2} = \frac{e^{2 a} + e^{- 2 a} + 2}{12} + \frac{e^{2 a} + e^{- 2 a} - 2}{4}$
$x^{2} + y^{2} = \frac{e^{2 a} + e^{- 2 a} + 2 + 3 e^{2 a} + 3 e^{- 2 a} - 6}{12}$
$x^{2} + y^{2} = \frac{4 e^{2 a} + 4 e^{- 2 a} - 4}{12}$
$x^{2} + y^{2} = \frac{e^{2 a} + e^{- 2 a} - 1}{3}$
$x^{2} + y^{2} = \frac{e^{2 a} + e^{- 2 a} + 2 - 3}{3}$
$x^{2} + y^{2} = \frac{{(e^{a} + e^{- a})}^{2}}{3} - 1$
$x^{2} + y^{2} = \frac{12 x^{2}}{3} - 1$
$x^{2} + y^{2} = 4 x^{2} - 1$
$3 x^{2} - y^{2} = 1 \Rightarrow \frac{x^{2}}{1 / 3} - y^{2} = 1$
eccentricity of hyperbola $= \sqrt{1 + \frac{b^{2}}{a^{2}}}$
$e = \sqrt{1 + \frac{1}{1 / 3}} = \sqrt{1 + 3}$
$e = 2$

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Eccentricity of the conic represented by x=ea+e−a2√3,y=ea−e−a2 is

Eccentricity of the conic represented by $x = \frac{e^{a} + e^{- a}}{2 \sqrt{3}}, y = \frac{e^{a} - e^{- a}}{2}$ is