Finding a tangent to an ellipse parallel to a given line Problem: Find the lines that are tangent to the ellipse $x^2 + 4y^2 = 8$ and parallel to $x +2y = 6$.
I tried to find the derivative of $x^2 + 4y^2 = 8$ and I got: $$\frac{dx}{dy} = -\frac{x}{2y}.$$
Not quite sure if it's right, but I tried to equate it with the slope of $x +2y = 6$, which is $-\frac{1}{2}$. So I get, $$-\frac{1}{2} = -\frac{x}{2y}$$ 
Now, I'm stuck and I don't know what to do. How do I get the tangent lines? I have no knowledge of any point of that tangent line, only the slope. I appreciate any help. Thanks.

Rewrite the equation of the ellipse $x^2+4y^2=8$ as
$$
\begin{align}
x^2+4y^2&=8\\
4y^2&=8-x^2\\
y^2&=\frac{8-x^2}{4}\\
y&=\left(\frac{8-x^2}{4}\right)^{\frac12}.\tag1
\end{align}
$$
The slope of the lines that are tangent to the ellipse is
$$
\begin{align}
\frac{dy}{dx}&=\frac{d}{dx}\left(\frac{8-x^2}{4}\right)^{\frac12}\\
&=\frac12\left(\frac{8-x^2}{4}\right)^{-\frac12}\left(\frac{-2x}{4}\right)\\
&=\frac12\left(\frac{4}{8-x^2}\right)^{\frac12}\left(-\frac{1}{2}x\right)\\
&=-\frac{1}{2}x\left(\frac{1}{8-x^2}\right)^{\frac12}
\end{align}
$$
or the quickest way
$$
\frac d{dx}(x^2+4y^2=8)\quad\Rightarrow\quad 2x+8y \frac {dy}{dx}=0\quad\Rightarrow\quad \frac {dy}{dx}=-\frac {x}{4y}.
$$
Since it is parallel with the line $x +2y = 6$, then
$$
\begin{align}
-\frac{1}{2}x\left(\frac{1}{8-x^2}\right)^{\frac12}&=-\frac{1}{2}\\
x\left(\frac{1}{8-x^2}\right)^{\frac12}&=1.\tag2
\end{align}
$$
Square both sides of $(2)$, yield
$$
\begin{align}
\frac{x^2}{8-x^2}&=1\\
x^2&=8-x^2\\
2x^2&=8\\
x_{1,2}&=\pm2\tag3
\end{align}
$$
$y$ can be found by plugging in $(3)$ to $(1)$, yield
$$
\begin{align}
y_{1,2}&=\left(\frac{8-2^2}{4}\right)^{\frac12}\\
&=\pm1,
\end{align}
$$
where $(x_{1,2},y_{1,2})$ are the points of contact. Now, the equation of tangent line can be found by using
$$
\frac{y-y_{1,2}}{x-x_{1,2}}=-\frac12.
$$
