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.

Not a synthetic proof, but ...
It's interesting to note that four target points are easy to identify: these are the corners of the rectangle that bounds the ellipse on the left, right, top, and bottom; call this the "axis-aligned bounding rectangle". Clearly, if we rotate the ellipse about its center, then the corners of the (new, un-rotated) axis-aligned bounding rectangle of the rotated figure also correspond to four target points. We can show that the target points form a circle by showing that these axis-aligned rectangles always have semi-diagonal $\sqrt{a^2 + b^2}$, regardless of rotation.
Parameterizing the ellipse by $P(a\cos\theta,b\sin\theta)$ and rotating about the center by angle $\phi$ gives an ellipse parameterized by
$$(a \cos\theta \cos\phi - b \sin\theta \sin\phi, a \cos\theta \sin\phi + b \sin\theta \cos\phi )$$
The right side of the axis-aligned bounding rectangle is determined by the value of $\theta$ that maximizes the $x$ coordinate; the top side is determined by the value that maximizes the $y$ coordinate.
Note that both parameterized coordinates have the form $U\cos\theta + V\sin\theta$, which can be expressed as $\sqrt{U^2+V^2} \sin\left(\theta+\psi\right)$ for some $\psi$. As $\theta$ traverses all values from $0$ to $2\pi$, the coordinate formula must attain a maximum value of $\sqrt{U^2 + V^2}$ (and we needn't worry about the exact value of $\psi$).
Consequently, the rotated ellipse's maximal $x$-coordinate is $x_{\star}:=\sqrt{a^2\cos^2\phi + b^2\sin^2\phi}$ and its maximal $y$-coordinate is $y_{\star}:=\sqrt{a^2\sin^2\phi + b^2\cos^2\phi}$, so that the rectangle's semi-diagonal, $d$, satisfies
$$d^2 = x_{\star}^2 + y_{\star}^2 = \left( a^2 + b^2 \right)\left( \cos^2\phi + \sin^2\phi \right) = a^2 + b^2$$
Therefore, the corners of the axis-aligned bounding rectangles of the rotated ellipse lie on the circle of radius $\sqrt{a^2 + b^2}$, regardless of the angle of rotation, $\phi$; rotating them back through angle $\phi$ won't change that fact.
This shows that all of the points with the "perpendicular tangents" property lie on a particular circle, but it does not show that all points on that circle have the "perpendicular tangents" property. That implication is left to the reader.
