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.
Geometry – Locus of Intersection Point of Two Perpendicular Tangents to an Ellipse
conic sectionsgeometry
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Kindly explain the reason behind this fact.
The reason is that an ellipse can be obtained by stretching/shrinking a circle. The strech/shrink is a linear map (linear transformation).
Let's consider two tangent lines on the circle $x^2+y^2=a^2$ at $(a\cos\theta,a\sin\theta)$,$(a\cos(\theta+\pi),a\sin(\theta+\pi))$. You already know that the two tangent lines are parallel.
Now, let's stretch/shrink the circle and the tangent lines. Stretching/shrinking the circle $x^2+y^2=a^2$ to obtain the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ means that you replace $y$ in $x^2+y^2=a^2$ with $\frac{a}{b}y$ to have $x^2+\left(\frac aby\right)^2=a^2$ which is nothing but $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
By this stretch/shrink, we have the followings :
The circle $x^2+y^2=a^2$ is transformed to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
The two parallel lines are transformed to two parallel lines.
The two lines tangent to the cirlce are transformed to two lines tangent to the ellipse.
The tangent points $(a\cos\theta,a\sin\theta)$,$(a\cos(\theta+\pi),a\sin(\theta+\pi))$ on the circle are transformed to two tangent points $(a\cos\theta,b\sin\theta)$,$(a\cos(\theta+\pi),b\sin(\theta+\pi))$ on the ellipse respectively.
From the above facts, it follows that the eccentric angles of the points of contact of two parallel tangents differ by $\pi$.
The followings are the proof for the above facts.
Let's consider the circle $x^2+y^2=a^2$ and two points $(a\cos\theta,a\sin\theta)$,$(a\cos(\theta+\pi),a\sin(\theta+\pi))$.
The equation of the tangent line at $(a\cos\theta,a\sin\theta)$ is given by $$a\cos\theta\ x+a\sin\theta\ y=a^2\tag1$$
Similarly, the equation of the tangent line at $(a\cos(\theta+\pi),a\sin(\theta+\pi))$ is given by $$a\cos(\theta+\pi)x+a\sin(\theta+\pi)y=a^2\tag2$$
Now, let's stretch/shrink the circle and the lines $(1)(2)$ by replacing $y$ with $\frac aby$ to have $$(1)\to a\cos\theta\ x+a\sin\theta\cdot\frac aby=a^2\tag3$$ $$(2)\to a\cos(\theta+\pi)x+a\sin(\theta+\pi)\cdot\frac aby=a^2\tag4 $$ Here note that these lines $(3)(4)$ are parallel since the slope of each line is $\frac{-b\cos\theta}{a\sin\theta}$.
Finally, note that $(3)$ can be written as $$\frac{a\cos\theta}{a^2}x+\frac{b\sin\theta}{b^2}y=1\tag5$$ which is nothing but the tangent line at $(a\cos\theta,b\sin\theta)$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
Similarly, $(4)$ can be written as $$\frac{a\cos(\theta+\pi)}{a^2}x+\frac{b\sin(\theta+\pi)}{b^2}y=1\tag6$$ which is nothing but the tangent line at $(a\cos(\theta+\pi),b\sin(\theta+\pi))$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
Since $(5)(6)$ are parallel, we see that the eccentric angles of the points of contact of two parallel tangents differ by $\pi$. $\quad\square$
The correct equation of the chord of contact is: $$ \frac{hx}{2}+ky-\frac {h^2+k^2}{3}=0 $$ and its distance from the origin (i.e. the radius of the tangent circle) is thus $$ r={2\over3}{h^2+k^2\over \sqrt{h^2+4k^2}}={2\over\sqrt3}{1\over \sqrt{1+k^2}}. $$ From that, it's easy to find minimum and maximum value of $r$.
Best Answer
It is a well known circle, generally called the "Director circle", though Wikipedia prefers "Fermat–Apollonius circle"
You can find a projective geometric proof here from Paris Pamfilos though I prefer one based on analytical geometry such as the discussion here by Michael Raugh or the following from slides 12 and 13 from Career Launcher India using earlier results for tangents to an ellipse: