Question

The eccentricity of the ellipse 25x² +16y² = 400 is​

Answer 1

$\textbf{Given:}$

$\mathsf{Ellipse\;is\;25x^2+16y^2=400}$

$\textbf{To find:}$

$\textsf{Eccentricity of the given ellipse}$

$\textbf{Solution:}$

$\textbf{Concept used:}$

$\boxed{\begin{minipage}{7cm} \mathsf{Eccentricity\;of\;the\;ellipse\;\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\;is}\\\\\mathsf{e=\sqrt{1-\dfrac{b^2}{a^2}}} \end{minipage}}$

$\mathsf{Consider,}$

$\mathsf{25x^2+16y^2=400}$

$\textsf{Divide bothsides by 400}$

$\mathsf{\dfrac{25x^2}{400}+\dfrac{16y^2}{400}=\dfrac{400}{400}}$

$\mathsf{\dfrac{x^2}{16}+\dfrac{y^2}{25}=1}$

$\mathsf{Here,\;a^2=25,\;b^2=16}$

$\mathsf{Then,\;Eccentricity}$

$\mathsf{e=\sqrt{1-\dfrac{b^2}{a^2}}}$

$\mathsf{e=\sqrt{1-\dfrac{16}{25}}}$

$\mathsf{e=\sqrt{\dfrac{25-16}{25}}}$

$\mathsf{e=\sqrt{\dfrac{29}{25}}}$

$\implies\boxed{\mathsf{e=\dfrac{3}{5}}}$

$\textbf{Find more:}$

The eccentricity of the ellipse represented by x = 5 (cos t + sin t), y = 3 (cos t – sin t) where t is parameter, is given by e, then 20e is equal to

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