Let the eccentricity of an ellipse x ^ 2 a ^ 2 + y ^ 2 b ^ 2 = 1, a...

Given,

$\dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1a > b$

Now using eccentricity formula, ${e}^{2} = 1 - \dfrac{{b}^{2}}{{a}^{2}}$

We get, $\dfrac{1}{16} = 1 - \dfrac{{b}^{2}}{{a}^{2}}$

$⇒ \dfrac{{b}^{2}}{{a}^{2}} = 1 - \dfrac{1}{16} = \dfrac{15}{16} ⇒ {b}^{2} = \dfrac{15}{16}{a}^{2}$

Now again $\dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1$ is passing through $\left(- 4\sqrt{\dfrac{2}{5}}, 3\right)$ on satisfying the point we get,

$⇒ \dfrac{16 \times \dfrac{2}{5}}{{a}^{2}} + \dfrac{9}{{b}^{2}} = 1$

$⇒ \dfrac{32}{5{a}^{2}} + \dfrac{9}{{b}^{2}} = 1$

Now putting the value ${b}^{2} = \dfrac{15}{16}{a}^{2}$ in above equation we get,

$\dfrac{32}{5{a}^{2}} + \dfrac{9}{\dfrac{15}{16}{a}^{2}} = 1$

$\dfrac{80}{5{a}^{2}} = 1$

$16 = {a}^{2}$

So, ${b}^{2} = 15 and {a}^{2} + {b}^{2} = 15 + 16 = 31$