Question

# The eccentricity of an ellipse $9{x^2} + 16{y^2} = 144$ is (a) $\dfrac{{\sqrt 3 }}{5}$(b) $\dfrac{{\sqrt 5 }}{3}$(c) $\dfrac{{\sqrt 7 }}{4}$(d) $\dfrac{2}{5}$

Hint: Here the eccentricity of an ellipse is a measure of how nearly circular is the ellipse. Eccentricity is found by the formula eccentricity = c/a where â€˜câ€™ is the distance from the centre to the focus of the ellipse and â€˜aâ€™ is the distance from the centre to the vertex for the standard form of the ellipse.

Given ellipse is $9{x^2} + 16{y^2} = 144$
Rewriting the ellipse, we get
$\dfrac{{9{x^2}}}{{144}} + \dfrac{{16{y^2}}}{{144}} = 1 \\ \\ \dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{9} = 1 \\$
For the ellipse of the form $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ the eccentricity is given by $e = \dfrac{c}{a}$, where $c = \sqrt {{a^2} - {b^2}}$.
Comparing both the equations we have $a = 4,{\text{ }}b = 3{\text{ }}$
So, $c = \sqrt {16 - 9} = \sqrt 7$
Therefore, $e = \dfrac{c}{a} = \dfrac{{\sqrt 7 }}{4}$
Thus, the answer is option (c) $\dfrac{{\sqrt 7 }}{4}$.

Note: The eccentricity of the ellipse is always greater than zero but less than one i.e. $0 < e < 1$. The standard form of the ellipse is $\dfrac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1$ with centre $\left( {h,k} \right)$. In this problem we have centre $\left( {0,0} \right)$ so we have used the ellipse form $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$.