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Q)

If $e$ is the eccentricity of the ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}\normalsize =1\;(a < b)$,then

$\begin{array}{1 1}b^2=a^2(1-e^2)\\a^2=b^2(1-e^2)\\a^2=b^2(e^2-1)\\b^2=a^2(e^2-1)\end{array}$

• In an ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1 if a< b,e is the eccentricity then,a^2=b^2(1-e^2) Answer : a^2=b^2(1-e^2) In an ellipse \large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$ if $a< b$ then,$a^2=b^2(1-e^2)$