Answer
We explain why $\overline {PQ} $ is perpendicular to the tangent to the ellipse at $Q$ is a consequence of the method of Lagrange multipliers.
Work Step by Step
Let the coordinates of $P$ and $Q$ be $P = \left( {{x_P},{y_P}} \right)$ and $Q = \left( {x,y} \right)$, respectively. Let $d$ be the distance from $P$ to $Q$. So,
$d = ||\overline {PQ} || = \sqrt {{{\left( {x - {x_P}} \right)}^2} + {{\left( {y - {y_P}} \right)}^2}} $
${\left( {x - {x_P}} \right)^2} + {\left( {y - {y_P}} \right)^2} = {d^2}$
Notice that this is the equation of a circle of radius $d$ centered at $P$.
Suppose the equation of an ellipse is given by $g\left( {x,y} \right) = 0$.
If we write $f\left( {x,y} \right) = {\left( {x - {x_P}} \right)^2} + {\left( {y - {y_P}} \right)^2}$, then the circles centered at $P$ are level curves of the function $f$. At the level curve $f\left( {x,y} \right) = {d^2}$ the circle touches the ellipse at $Q$ as is shown in Figure 17. Thus, finding the minimum distance $d$ is the same as minimizing the function $f$ on the ellipse $g\left( {x,y} \right) = 0$.
According to the method of Lagrange multipliers,
$\nabla f = \lambda \nabla g$
Since $\lambda$ is a scalar, this implies that $\nabla f$ is parallel to $\nabla g$. But the gradient $\nabla g$ is perpendicular to the tangent to the ellipse at $Q$. Thus, it follows that $\nabla f$ is also perpendicular to the tangent to the ellipse at $Q$. Since $\nabla f$ is perpendicular to the tangent to the circle at $Q$, so $\nabla f$ is parallel to $\overline {PQ} $. Thus, we conclude that $\overline {PQ} $ is also perpendicular to the tangent to the ellipse at $Q$. Therefore, this conclusion is a consequence of the method of Lagrange multipliers.
