Answer
The foci are ${F_1} = \left( {0,\sqrt {\frac{{12}}{5}} } \right)$ and ${F_2} = \left( {0, - \sqrt {\frac{{12}}{5}} } \right)$.
The focal vertices are $\left( {0, \pm b} \right) = \left( {0, \pm \frac{4}{{\sqrt 5 }}} \right)$ and the minor vertices are $\left( { \pm a,0} \right) = \left( { \pm \frac{2}{{\sqrt 5 }},0} \right)$.
Work Step by Step
We have
${\left( {2x + \frac{1}{2}y} \right)^2} = 4 - {\left( {x - y} \right)^2}$
Multiply both sides by $4$ gives
$4{\left( {2x + \frac{1}{2}y} \right)^2} = 16 - 4{\left( {x - y} \right)^2}$
${\left( {4x + y} \right)^2} = 16 - 4{\left( {x - y} \right)^2}$
${\left( {4x + y} \right)^2} + 4{\left( {x - y} \right)^2} = 16$
$16{x^2} + 8xy + {y^2} + 4\left( {{x^2} - 2xy + {y^2}} \right) = 16$
$16{x^2} + 8xy + {y^2} + 4{x^2} - 8xy + 4{y^2} = 16$
$20{x^2} + 5{y^2} = 16$
$\frac{5}{4}{x^2} + \frac{5}{{16}}{y^2} = 1$
${\left( {\frac{x}{{2/\sqrt 5 }}} \right)^2} + {\left( {\frac{y}{{4/\sqrt 5 }}} \right)^2} = 1$
By Theorem 1 of Section 12.5, this is the equation of an ellipse in standard position, where $a = \frac{2}{{\sqrt 5 }}$ and $b = \frac{4}{{\sqrt 5 }}$. Since $b>a>0$, we have
$c = \sqrt {\frac{{16}}{5} - \frac{4}{5}} = \sqrt {\frac{{12}}{5}} $
The foci are ${F_1} = \left( {0,\sqrt {\frac{{12}}{5}} } \right)$ and ${F_2} = \left( {0, - \sqrt {\frac{{12}}{5}} } \right)$.
The focal vertices are $\left( {0, \pm b} \right) = \left( {0, \pm \frac{4}{{\sqrt 5 }}} \right)$ and the minor vertices are $\left( { \pm a,0} \right) = \left( { \pm \frac{2}{{\sqrt 5 }},0} \right)$.
