How to find the point on an ellipse that is closest to the point A outside of the ellipse I am working on a project where I need a formula to find the point $B(x,y)$ on the ellipse $x^2+4y^2=r^2$ that is closest to the point $A(x_0,y_0)$, where $A$ is known and is outside the ellipse. 
I have been trying to use the distance formula $d=\sqrt{(x-x_0)^2+(y-y_0)^2}$ but I have been running into some problems trying differentiate.
Any help would be greatly appreciated.
 A: Not sure if efficient, but here is how I would do it: 
Rewrite the ellipse as a function: $$f_{1,2}(x):=\pm\frac{1}{2} \sqrt{r^2-x^2}$$ 
Using the distance formula, we get the following functions: $$d_{1,2}(x):=\sqrt{(f_{1,2}(x)-y_0)^2+(x-x_0)^2}$$ 

Now we minimize $d_{1,2}(x)$ by solving ${d_{1,2}}'(x)=0$: 
Suppose $A$ is $(1,1)$ and $r=1$, so we use $f_1(x)$: $$d_1(x)=\sqrt{\left(\frac{\sqrt{1-x^2}}{2}-1\right)^2+(x-1)^2}\\d_1'(x)=\frac{x \left(\frac{1}{\sqrt{1-x^2}}+\frac{3}{2}\right)-2}{\sqrt{3 x^2-4 \sqrt{1-x^2}-8 x+9}}=0\tag{1}$$ 
Which gives one solution: $$\{\{x\to 0.69282\}\}$$ 
Plug this in $f_1(x)$ and we get $B(x,y)$ as: $$B(x,y)=\{0.69282,0.360555\}$$

If you have trouble solving for $x$ in$(1)$, you can always use numerical methods like Newton-Raphson.
A: 
(see Desmos implementation here.)
New Answer
Point on Ellipse:
$$B(x,y)=(r\cos\theta, \frac r2\sin\theta)\\
B'(x', y')=(-r\sin\theta, \frac r2 \cos\theta)\\$$
When $B$ is the nearest point to $A$, 
$$\begin{align}
(\mathbf b-\mathbf a)\cdot \mathbf b'
&=0\\
\left(\begin{array}.r\cos\theta-x_0\\\frac r2\sin\theta-y_0\end{array}\right)
\cdot \left(-r\sin\theta\atop \frac r2\cos\theta\right)
&=0\\
(r\cos\theta-x_0)(-r\sin\theta)+(\tfrac r2\sin\theta-y_0)(\tfrac r2\cos\theta)
&=0\\
-2r\sin\theta+2x_0\tan\theta+\frac r2\sin\theta-y_0&=0\\
\tfrac 32 r \sin\theta-2x_0\tan\theta+y_0&=0
\end{align}$$
Solving for $\theta$ gives coordinates for $B$. 

Previous Answer
Point on Ellipse:
$$B(x,y)=(r\cos\theta, \frac r2\sin\theta)\\
B'(x', y')=(-r\sin\theta, \frac r2 \cos\theta)\\
\frac {dy}{dy}=-\frac 1{2\tan\theta}=m_1\qquad \text{(slope at $B$)}$$
Slope of $AP$:
$$m_2=\frac {y-\frac r2\sin\theta}{x_0-r\cos\theta}$$
$B$ is the nearest point on the ellipse to $A$ when $AB$ is perpendicular to the tangent at $B$, i.e.
$$m_1m_2=-1\\
-\frac 1{2\tan\theta}\cdot\frac {y-\frac r2\sin\theta}{x_0-r\cos\theta}=-1\\
2x_0\tan\theta-\frac 32r\sin\theta=y_0$$
Solving for $\theta$ gives the coordinates for $B$. 
