The locus of the intersection point of two perpendicular tangents to a given ellipse 
For a given ellipse, find the locus of all points P for which the two tangents are perpendicular.

I have a trigonometric proof that the locus is a circle, but I'd like a pure (synthetic) geometry proof.

If all you want is a proof that the locus is a circle, we may assume that the ellipse is given by
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$$
Ignoring vertical tangents for now, 
if a line $y=mx+k$ is tangent to the ellipse, then plugging in this value of $y$ into the equation for the ellipse gives
$$\frac{x^2}{a^2} + \frac{(m^2x^2 + 2mkx + k^2)}{b^2} = 1$$
or
$$(b^2 + a^2m^2)x^2 + 2a^2mkx + (a^2k^2 - a^2b^2) = 0.$$
This equation gives the two points of intersection of the line with the ellipse. If the line is tangent, then the two points must coincide, so the quadratic must have zero discriminant. That is, we need
$$(2a^2mk)^2 - 4(a^2k^2 - a^2b^2)(b^2+a^2m^2) = 0$$
or equivalently, 
$$\begin{align*}
(a^2m^2)k^2 -a^2(b^2+a^2m^2)k^2 &= -a^2b^2(b^2+a^2m^2)\\
-a^2b^2k^2&= -a^2b^2(b^2+a^2m^2)\\
k^2 &= b^2+a^2m^2\\
k &= \pm\sqrt{a^2m^2 + b^2}.
\end{align*}
$$
So the lines that are tangent to the ellipse are of the form
$$y = mx \pm \sqrt{a^2m^2 + b^2}.$$
Since the problem is symmetric about $x$ and $y$, consider the points on the upper half plane, so that we will take the plus sign above. The tangent perpendicular to this one will therefore have equation
$$y = -\frac{1}{m}x + \sqrt{\frac{a^2}{m^2} + b^2},$$
or equivalently
$$my = -x + \sqrt{a^2 + m^2b^2}.$$
(We are ignoring the vertical and horizontal tangents; I'll deal with them at the end).
If a point $(r,s)$ is on both lines, then we have
$$\begin{align*}
s-mr &= \sqrt{a^2m^2 + b^2}\\
ms + r &= \sqrt{a^2+m^2b^2}.
\end{align*}$$
Squaring both sides of both equations we get
$$\begin{align*}
s^2 - 2mrs + m^2r^2 &= a^2m^2 + b^2\\
m^2s^2 + 2mrs + r^2 &= a^2 + m^2b^2
\end{align*}$$
and adding both equations, we have
$$\begin{align*}
(1+m^2)s^2 + (1+m^2)r^2 &= (1+m^2)a^2 + (1+m^2)b^2,\\
(1+m^2)(s^2+r^2) &= (1+m^2)(a^2+b^2)\\
s^2+r^2 = a^2+b^2,
\end{align*}$$
showing that $(s,r)$ lies in a circle, namely $x^2+y^2 = a^2+b^2$. 
Taking the negative sign for the square root leads to the same equation.
Finally, for the vertical and horizontal tangents, these occur at $x=\pm a$; the horizontal tangents are $y=\pm b$. Their intersections occur at $(\pm a,\pm b)$, which lie on the circle given above. So the locus of such points is contained in the circle $x^2+y^2 = a^2+b^2$. 
Conversely, consider a point $(r,s)$ that lies on $x^2+y^2 = a^2+b^2$. If a tangent to the ellipse
$$ y = mx + \sqrt{a^2m^2 + b^2}$$
goes through $(r,s)$, then we have
$$ s-mr = \sqrt{a^2m^2+b^2}.$$
Squaring both sides, we have
$$s^2 - 2msr + m^2r^2 = a^2m^2 + b^2$$
or
$$(a^2-r^2)m^2 +2srm + (b^2-s^2) = 0.$$
Since $r^s+s^2 = a^2+b^2$, then $a^2 - r^2 = s^2-b^2$, we we have
$$(s^2-b^2)m^2 + 2srm + (b^2-s^2) = 0,$$
and if we do not have $s=\pm b$ (the horizontal/vertical tangent intersection points), then we get
$$m^2 + tm - 1 = 0,\qquad\text{where } t = \frac{2sr}{s^2-b^2}.$$
So the two solutions for $m$, $m_1$ and $m_2$, satisfy $m_1m_2 = -1$, hence the two tangents are perpendicular. That is, at every point on the (upper half of the) circle, the two lines through the point that are tangent to the ellipse are perpendicular to each other.
So all such points are on the circle, and all points on the circle are such points. (The circle is called the director circle of the ellipse).
