The eccentricity of the conic {x^2} - 4{y^2} | vedclass

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Question

(d) Given conic is $\frac{{{x^2}}}{{{{(1)}^2}}} - \frac{{{y^2}}}{{{{\left( {\frac{1}{2}} ight)}^2}}} = 1$

$	herefore {b^2} = {a^2}({e^2} - 1)$

Vedclass · Accepted Answer

$\frac{{\sqrt 5 }}{2}$

Vedclass · Answer

(d) Given conic is $\frac{{{x^2}}}{{{{(1)}^2}}} - \frac{{{y^2}}}{{{{\left( {\frac{1}{2}} ight)}^2}}} = 1$

$	herefore {b^2} = {a^2}({e^2} - 1)$

$&rArr;$ $\frac{1}{4} + 1 = {e^2}$

$&rArr;$  $e = \frac{{\sqrt 5 }}{2}$.

The eccentricity of the conic ${x^2} - 4{y^2} = 1$, is

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