Question

# The eccentricity of the ellipse $$9{x^2} + 16{y^2} = 576$$ is

A
72
B
54
C
712
D
74

Solution

## The correct option is D $${{\sqrt 7 } \over 4}$$Consider the given ellipse. $$9{{x}^{2}}+16{{y}^{2}}=576$$ $$\dfrac{9{{x}^{2}}}{576}+\dfrac{16{{y}^{2}}}{576}=1$$ $$\dfrac{{{x}^{2}}}{64}+\dfrac{{{y}^{2}}}{36}=1$$ $$\dfrac{{{x}^{2}}}{{{8}^{2}}}+\dfrac{{{y}^{2}}}{{{6}^{2}}}=1$$   We know that the general equation of the ellipse is, $$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$$   On comparing, we get $$a=8,b=6$$   We know that the eccentricity of the ellipse, $$e=\sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}}$$ $$e=\sqrt{1-\dfrac{36}{64}}$$ $$e=\sqrt{\dfrac{28}{64}}$$ $$e=\dfrac{\sqrt{7}}{4}$$ So, the value of the eccentricity is $$\dfrac{\sqrt{7}}{4}$$.Mathematics

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