Problem 718

Sir SagarMay 14, 2015May 14, 20150

Find the eccentricity of the ellipse $\frac{x^{2}}{25}+\frac{(y-7)^{2}}{5}=1$

Solution:-

eccentricity of the ellipse = $\frac{c}{a}$

c = $\sqrt{a^{2}-b^{2}}$

c = $\sqrt{25-5}$

c= $\sqrt{20}$

c = $2\sqrt{5}$

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

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