Let an ellipse with centre (1, 0) and latus rectum of length...

Given,

An ellipse with centre $\left(1, 0\right)$ and latus rectum of length $\dfrac{1}{2}$ have its major axis along x-axis

And its minor axis subtends an angle ${60}^{∘}$ at the foci,

So, on plotting the diagram we get,

Now from diagram we get,

$\dfrac{2{b}^{2}}{a} = \dfrac{1}{2} ⇒ {b}^{2} = \dfrac{a}{4} . . . \left(1\right)$

And $tan{30}^{∘} = \dfrac{b}{ae}$

$⇒ {\left(\dfrac{1}{\sqrt{3}}\right)}^{2} = {\left(\dfrac{b}{ae}\right)}^{2}$

$⇒ \dfrac{1}{3} = \dfrac{{b}^{2}}{{a}^{2} - {b}^{2}}$

$⇒ {a}^{2} - {b}^{2} = 3{b}^{2}$

$⇒ {b}^{2} = \dfrac{{a}^{2}}{4} . . . . . . . \left(2\right)$

Now from equation \left(1\right) \text{ and } \left(2\right)$we get,</p><p>$⇒ a = 1, {b}^{2} = \dfrac{1}{4} ⇒ b = \dfrac{1}{2}$</p><p>Hence,${\left(2a + 2b\right)}^{2} = 9