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$x = \verts{a}\cos\pars{\theta}$ and $y = \verts{b}\sin\pars{\theta}$ satisfy the ellipse equation. So
$$
m\braces{\verts{a}\cos\pars{\theta}} + n\braces{\verts{b}\sin\pars{\theta}} = k
\quad\imp\quad
\sin\pars{\theta} + {m\verts{a} \over n\verts{b}}\,\cos\pars{\theta}
= {k \over n\verts{b}}
$$
Let's $\ds{\tan\pars{\mu} = {m\verts{a} \over n\verts{b}}}$:
$$\!\!\!\!\!
\sin\pars{\theta}\cos\pars{\mu} + \cos\pars{\theta}\sin\pars{\mu}
= {k \over n\verts{b}}\,\cos\pars{\mu}
= {k \over n\verts{b}}\,{1 \over \root{\tan^{2}\pars{\mu} + 1}}
={k\sgn\pars{n} \over \root{m^{2}a^{2} + n^{2}b^{2}}}
$$
$$
\sin\pars{\theta + \mu} = {k\sgn\pars{n} \over \root{m^{2}a^{2} + n^{2}b^{2}}}
$$
If $\ds{{\verts{k} \over \root{m^{2}a^{2} + n^{2}b^{2}}} \leq 1}$ we have solutions for $\theta\ \in\ {\mathbb R}$ $\pars{~\mbox{otherwise,}\ \theta\ \in\ {\mathbb C}~}$:
$$
\theta = -\arctan\pars{m\verts{a} \over n\verts{b}} + \arcsin\pars{{k\sgn\pars{n} \over \root{m^{2}a^{2} + n^{2}b^{2}}}}
+ 2\pi n\,,\quad n\ \in\ {\mathbb Z}
$$
where $0 \leq \arcsin\pars{\alpha} < 2\pi$
Once we know $\theta$ we can evaluate $x = \verts{a}\cos\pars{\theta}$ and
$y = \verts{b}\sin\pars{\theta}$:
\begin{align}
x&=\verts{a}\cos\pars{\arctan\pars{m\verts{a} \over n\verts{b}}}
\cos\pars{\arcsin\pars{k\sgn\pars{n} \over \root{m^{2}a^{2} + n^{2}b^{2}}}}
\\[3mm]&+
\verts{a}\sin\pars{\arctan\pars{m\verts{a} \over n\verts{b}}}
\sin\pars{\arcsin\pars{k\sgn\pars{n} \over \root{m^{2}a^{2} + n^{2}b^{2}}}}
\\[3mm]&=
\verts{a}\,{1 \over \root{\pars{ma/nb}^{2} + 1}}\,
\root{1 - {k^{2} \over m^{2}a^{2} + n^{2}b^{2}}}
\\[3mm]&+
\verts{a}\,{m\verts{a}/\pars{n\verts{b}} \over \root{\pars{ma/nb}^{2} + 1}}\,
{k\sgn\pars{n} \over \root{m^{2}a^{2} + n^{2}b^{2}}}
\end{align}
$$
\color{#00f}{\large%
\begin{array}{rcl}
x & = &
{\verts{nba}\sqrt{m^{2}a^{2} + n^{2}b^{2} - k^{2}} + ma^{2}k \over m^{2}a^{2} + n^{2}b^{2}}
\\[3mm]
y & = & {k \over n} - {m \over n}\,x
\\[3mm]&&
\color{#000}{\large\mbox{when}\
{\verts{k} \over \root{m^{2}a^{2} + n^{2}b^{2}}} \leq 1}
\end{array}}
$$
