The problem, presented in a previous post, of determining the point on a circumference that is closest to a given interior point \(P\), changes radically if we consider an ellipse, instead of a circle, even if we limit ourself to the case where the point \(P\) lies on the major axis.

Consider the ellipse with equation

\[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\]

with \(a>b\), and a point \(P\) inside it with coordinates \(P(x_P,0)\), with \(x_P\leq a\). We wish to find the points on the ellipse that have minimum distance from \(P\), and the amount of such distance.

The locus of points having a given distance from \(P\) is a circumference centered in \(P\). So, we should look for the larger circle entirely contained inside the ellipse, i.e. tangent to the ellipse. The tangent points \(Q_1\) and \(Q_2\) are the points we are looking for, as depicted in the Figure below.

Let us consider the circumference with center in \(P\) and radius \(r\), whose equation is

\[(x-x_P)^2+y^2 = r^2.\]

The intersection points between circle and ellipse are found by solving the system of equations

\begin{equation}\begin{cases}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\

(x-x_P)^2+y^2 = r^2. \end{cases}\tag{1}\label{eq578:1}\end{equation}

Determining \(y^2\) from the second equation and plugging it into the first one leads to the second order equation

\begin{equation}x^2\left(\frac{1}{b^2}-\frac{1}{a^2}\right) -\frac{2x_P}{b^2}x-\frac{r^2-x_P^2-b^2}{b^2} = 0.\tag{2}\label{eq578:2}\end{equation}

For the desired radius, the circumference meets the ellipse at only two points *having the same abscissa.* Thus the above equation must have **only one valid solution**, which necessarily occurs if its discriminant is equal to zero, meaning

\[\frac{x_P^2}{b^4}+\left(\frac{1}{b^2}-\frac{1}{a^2}\right)\frac{r^2-x_P^2-b^2}{b^2} = 0\]

Solving with respect to \(r\) gives

\begin{equation}r^2 =b^2 -\frac{b^2}{a^2-b^2}x_P^2.\tag{3}\label{eq578:3}\end{equation}

For this value of \(r\), equation \eqref{eq578:2} becomes a perfect square, having the only solution

\[x = x_P \cdot \frac{a^2}{a^2-b^2}.\]

Replacing this value of \(x\) into the circle equation yields the coordinates of \(Q_1\) and \(Q_2\).

As an exercise, you can test the correctness of this procedure with the ellipse of major semi-axis \(a=4\) and minor semi-axis \(b=2\). If \(P\) has coordinates \((1,0)\) you should find \(r=\sqrt{\frac{11}{3}}\),

\[Q_1\left(\frac{4}{3},

\frac{4\sqrt{2}}{3} \right) \]

and

\[Q_2\left(\frac{4}{3},

-\frac{4\sqrt{2}}{3} \right).\]

However, discriminant of \eqref{eq578:2} being null is only a *sufficient condition* for the system \eqref{eq578:1} to have only one valid solution. Why? Try to see what happens with the same ellipse and point \(P(\frac{7}{2},0)\). Check that in this case the closest point to \(P\) on the ellipse is \(A(a,0)\), and thus

\begin{equation}r = a-x_P.\tag{4}\label{eq578:4}\end{equation}

What is the range of values of \(x_P\) allowing you to use \eqref{eq578:3}? And when do you need to use \eqref{eq578:4} instead?