Let an ellipse with centre $$(1, 0)$$ and latus rectum of length $$\frac{1}{2}$$ have its major axis along x-axis. If its minor axis subtends an angle $$60°$$ at the foci, then the square of the sum of the lengths of its minor and major axes is equal to _____.

Given $$LR = \frac{2b^2}{a} = \frac{1}{2} \implies b^2 = \frac{a}{4}$$

Angle subtended at foci is $$60^\circ \implies \theta = 60^\circ \implies \theta/2 = 30^\circ$$

$$\tan(30^\circ) = \frac{b}{ae} \implies \frac{1}{\sqrt{3}} = \frac{b}{ae} \implies ae = b\sqrt{3}$$

Substitute $$e^2 = 1 - \frac{b^2}{a^2}$$ into $$a^2e^2 = 3b^2$$:

$$a^2\left(1 - \frac{b^2}{a^2}\right) = 3b^2 \implies a^2 - b^2 = 3b^2 \implies a^2 = 4b^2$$

Use $$b^2 = \frac{a}{4}$$:

$$a^2 = 4\left(\frac{a}{4}\right) = a \implies a = 1 \text{ (since } a > 0\text{)}$$

$$b^2 = \frac{1}{4} \implies b = \frac{1}{2}$$

Length of major axis $$= 2a = 2$$; Length of minor axis $$= 2b = 1$$

Sum of axes $$= 2 + 1 = 3$$

Square of the sum $$= 3^2 = 9$$