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

We are given an ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ with $$a > b$$ and eccentricity $$e = \frac{1}{4}$$. For such an ellipse, the eccentricity satisfies

$$e^2 = 1 - \frac{b^2}{a^2}$$

so

$$\frac{1}{16} = 1 - \frac{b^2}{a^2},$$

which implies

$$\frac{b^2}{a^2} = 1 - \frac{1}{16} = \frac{15}{16},$$

and hence

$$b^2 = \frac{15a^2}{16}.$$

The ellipse also passes through the point $$\left(-4\sqrt{\frac{2}{5}},\;3\right)$$. Substituting this into the equation of the ellipse gives

$$\frac{\left(-4\sqrt{\frac{2}{5}}\right)^2}{a^2} + \frac{3^2}{b^2} = 1,$$

which simplifies to

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

and therefore

$$\frac{32}{5a^2} + \frac{9}{b^2} = 1.$$

Next, substituting $$b^2 = \frac{15a^2}{16}$$ into this equation yields

$$\frac{32}{5a^2} + \frac{9 \times 16}{15a^2} = 1,$$

or

$$\frac{32}{5a^2} + \frac{144}{15a^2} = 1.$$

Since $$\frac{144}{15} = \frac{48}{5},$$ this becomes

$$\frac{32}{5a^2} + \frac{48}{5a^2} = 1,$$

so

$$\frac{80}{5a^2} = 1 \quad\Longrightarrow\quad \frac{16}{a^2} = 1 \quad\Longrightarrow\quad a^2 = 16.$$

Substituting back to find $$b^2$$ gives

$$b^2 = \frac{15 \times 16}{16} = 15.$$

Finally,

$$a^2 + b^2 = 16 + 15 = 31.$$

The correct answer is Option B: $$31$$.