Tricky Chain Rule/Differentation/Derivative/Ellipse problem 
When I found the derivative I got:
$$\frac{1}{2}x+ 2y\frac{dy}{dx}=0$$
$$\frac{dy}{dx}=\frac{\frac{-1}{2}x}{2y}$$
$$\frac{dy}{dx}=\frac{-1x}{4y}$$
But I cannot plug in $(4, 0)$ and since point $Q$ is NOT in the circle equation.
By the way the "solution" is $$y = \frac{-x}{2\sqrt{3}} + \frac{2}{\sqrt{3}}$$
But I don't know how that is the solution? Since its from the answer key without steps shown
I mean I could guess and check the slope and plug in any value in the circle equation:
$$y = \pm \sqrt{1 - \frac{1}{4}x^{2}}$$
since the domain of $x$ is $(-2, 2)$.
But this is obviously not how you solve it.emphasized text
 A: Notice, let point of tangency $P(h, k)$ on the ellipse hence it will satisfy the equation of the ellipse $\frac{x^2}{4}+y=1$ as follows $$\frac{h^2}{4}+k^2=1\tag 1$$
Now, the slope of the tangent at the point $(h, k)$ is $$\left(\frac{dy}{dx}\right)_{(h, k)}=\text{slope of line PQ joining the points}\ (h, k) \ \text{&}\ (4, 0) $$ $$\frac{-h}{4k}=\frac{k-0}{h-4}$$ 
$$-h^2+4h=4k^2\iff \frac{h^2}{4}+k^2=h\tag 2$$
Now, comparing (1) & (2), we get 
$$h=1$$
Hence, setting this value of $h$ in (1), we get $$\frac{(1)^2}{4}+k^2=1$$ $$k^2=\frac{3}{4}\iff k=\pm\frac{\sqrt 3}{2}$$ 
Since, the point $P$ is in the first quadrant its y-coordinate $k$ will be positive i.e. $k=\frac{\sqrt 3}{2}$ so we have $P\left(1, \frac{\sqrt3}{2}\right)$
Hence, the equation of the tangent PQ joining the points $P\left(1, \frac{\sqrt3}{2}\right)$ & $Q(4, 0)$ is given as $$y-0=\frac{\frac{\sqrt3}{2}-0}{1-4}(x-4)$$
$$\color{red}{x+2y\sqrt 3-4=0}$$
A: Differentiate using Chain Rule getting equation of tangent ( some steps omitted )
$$ \frac{x x_P}{4} + \frac{y y_P}{1} =1  \tag{1} $$
You are given
$$ x=x_Q=4,y=y_Q=0, \rightarrow x_P= 1 \tag{2}$$
Substitute into equation  of ellipse, you get $ y_P= \frac{\sqrt3}{2} \tag{3}$
Tangent line has equation:
$$\frac{x\cdot 1}{4} + \frac {y \sqrt 3}{2} =1. \tag{4}$$
