Given real numbers $x$ and $y$ that satisfy
$$\frac { x ^ { 2 } } { 2 } + \frac { y ^ { 2 } } { 4 } = 1 , \quad x \geqq 0 , \quad y \geqq 0$$
we are to find the maximum value of
$$P = x ^ { 2 } + x y + y ^ { 2 } .$$
Let $x$ and $y$ satisfy the conditions. When we set $x = \sqrt { 2 } \cos \theta \left( 0 \leqq \theta \leqq \frac { \pi } { 2 } \right)$, we have
$$y = \mathbf { A } \sin \theta .$$
Thus $P$ can be represented as
$$\begin{aligned}
P & = \sqrt { \mathbf { B } } \sin 2 \theta - \cos 2 \theta + \mathbf { C } \\
& = \sqrt { \mathbf { D } } \sin ( 2 \theta - \alpha ) + \mathbf { E }
\end{aligned}$$
where
$$\sin \alpha = \frac { \sqrt { \mathbf { F } } } { \mathbf { G } } , \quad \cos \alpha = \frac { \sqrt { \mathbf { H } } } { \mathbf { I } } \quad \left( 0 < \alpha < \frac { \pi } { 2 } \right) .$$
Hence the maximum value of $P$ is $\sqrt { \square \mathbf { J } } + \mathbf { K }$.\\
Let us denote the $\theta$ at which the value of $P$ is maximized by $\theta _ { 0 }$. Then we have
$$2 \theta _ { 0 } = \alpha + \frac { \pi } { \square } ,$$
and hence
$$\sin 2 \theta _ { 0 } = \frac { \sqrt { \mathbf { M } } } { \mathbf { N } } , \quad \cos 2 \theta _ { 0 } = - \frac { \sqrt { \mathbf { O } } } { \mathbf { N } } .$$