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) $\dfrac{2}{5}$

Answer 1

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 ‘c’ is the distance from the centre to the focus of the ellipse and ‘a’ 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 $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$.