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

Equation of ellipse is  $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1,\mathrm{a}>\mathrm{b}.$
Given eccentricity is $\frac{1}{4}\Rightarrow {\mathrm{e}}^2=1-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}$
$\frac{1}{16}=1-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}\Rightarrow \frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=1-\frac{1}{16}=\frac{15}{16}\Rightarrow {\mathrm{b}}^2=\frac{15}{16}{\mathrm{a}}^2$
Put the value of b² in equation of ellipse and satisfy the point 
$\begin{array}{l}\left(-4\sqrt{\frac{2}{5}},3\right)\Rightarrow \frac{16\times \frac{2}{5}}{{\mathrm{a}}^2}+\frac{9}{{\mathrm{b}}^2}=1\\ \Rightarrow \frac{32}{5{\mathrm{a}}^2}+\frac{9}{{\mathrm{b}}^2}=1\Rightarrow \frac{32}{5{\mathrm{a}}^2}+\frac{9}{\frac{15}{16}{\mathrm{a}}^2}=1\Rightarrow \frac{80}{5{\mathrm{a}}^2}=1\\ \Rightarrow 16={\mathrm{a}}^2\end{array}$
Put the value of a² in ${\mathrm{b}}^2\Rightarrow {\mathrm{b}}^2=\frac{15}{16}{\mathrm{a}}^2$
b² =15