The eccentricity of an ellipse having centre | vedclass

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Question

Centre at $(0,0)$

$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$

at point $(4,-1)$

$\frac{{16}}{{{a^2}}} + \frac{1}{{{b^2}}} = 1$

Vedclass · Accepted Answer

$\frac{{\sqrt 3 }}{2}$

Vedclass · Answer

Centre at $(0,0)$

$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$

at point $(4,-1)$

$\frac{{16}}{{{a^2}}} + \frac{1}{{{b^2}}} = 1$

$16{b^2} + {a^2} = {a^2}{b^2}\,\,\,\,\,\,\,\,.......\left( i \right)$ at point $(-2,2)$

$\frac{4}{{{a^2}}} + \frac{4}{{{b^2}}} = 1$

$4{b^2} + {a^2} = {a^2}{b^2}\,\,\,\,\,\,\,\,.......\left( {ii} \right)$

$ \Rightarrow 16{b^2} + {a^2} = 4{b^2} + {a^2}$

From equation $(i)$ and $(ii)$

$ \Rightarrow 3{a^2} = 12{b^2}$

$ \Rightarrow \boxed{{a^2} = 4{b^2}}$

${b^2} = {a^2}\left( {1 - {e^2}} \right)$

${e^2} = \frac{3}{4}$

$e = \frac{{\sqrt 3 }}{2}$

The eccentricity of an ellipse having centre at the origin, axes along the co-ordinate axes and passing through the points $(4,-1)$ and $(-2, 2)$ is

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