Let the eccentricity of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$, be $\frac{1}{4}$. If this ellipse passes through the point $\left(-4\sqrt{\frac{2}{5}}, 3\right)$, then $a^2+b^2$ is equal to :

$\begin{array}{l}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\mathrm{a}>\mathrm{b}\\ \\ e^2=1-\frac{b^2}{a^2}\\ \\ \frac{1}{16}=1-\frac{b^2}{a^2}\\ \\ \frac{b^2}{a^2}=1-\frac{1}{16}=\frac{15}{16} \Rightarrow b^2=\frac{15}{16}{a}^2\\ \\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ \\ \frac{16\times \frac{2}{5}}{a^2}+\frac{9}{b^2}=1\\ \\ \frac{32}{5a^2}+\frac{9}{b^2}=1\\ \\ \begin{array}{l}\frac{32}{5a^2}+\frac{9}{\frac{15}{16}{a}^2}=1\\ \\ \frac{80}{5a^2}=1\\ \\ 16=a^2\\ \\ b^2=15\end{array}\end{array}$