A triangle is circumscribed to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and two of its vertices lie on the directrices, such that one lies on each directrix. Then, the locus of the third vertex of the triangle is?
I tried using an affine transform to simplify the situation and began working with $x^2+y^2=1$, with one vertex lying on $x=1/e$ and the other on $x=-1/e$. Then I tried various approaches.
First, I tried writing the equations to the tangents and computing the third vertex, but it became extremely calculative and unpalatable.
So, I switched to a different approach, by finding the pole with respect to the lines $x=1/e$ and $x=-1/e$, which came out to be $(e,0)$ and $(-e,0)$ respectively. Then I tried to find some useful relations for chords passing through these two points, but all I could come up with was that the products of the two chord segments for a chord passing through either pole is constant (equal to $1-e^2$, a trivial result).
I am not able to figure out how to best simplify this problem. What would the most straightforward way to proceed be, without indulging in lengthy calculations?
First thing to do is to scale the ellipse, and its directrix lines by a factor of $(1/a)$ horizontally, and $(1/b)$ vertically. This will turn the ellipse into a unit circle, and will place the lines at $x = \pm \dfrac{1}{e} $.
In this transformed figure, if you have a tangent line whose direction vector is $ u = (\cos \theta , \sin \theta) $, then its normal vector will be $ n = (-\sin \theta, \cos \theta) $. Tangency will occur at $ r_1 = n $, and the equation of the tangent is
$ p(t) = n + t u = (- \sin \theta + t \cos \theta, \cos \theta + t \sin \theta) $
Vertices $A$ and $B$ as shown in the figure below, are determined by setting $ x = \dfrac{1}{e} $ and $ x = - \dfrac{1}{e} $. This gives
$ A = ( \dfrac{1}{e} , \dfrac{1}{\cos \theta} (1 + \sin \theta /e ) )$
$ B = ( - \dfrac{1}{e} , \dfrac{1}{\cos \theta} (1 - \sin \theta / e )) $
And we have the distances $c_1$ between $r_1$ and $A$, and the distance $c_2$ between $r_1$ and $B$ given by
$ c_1 = \dfrac{1}{\cos \theta} ( \dfrac{1}{e} + \sin \theta ) $
$ c_2 = \dfrac{1}{\cos \theta} ( \dfrac{1}{e} - \sin \theta ) $
Note that our circle is the incircle of the $\triangle ABC$, therefore, its center (which is the origin) satisfies
$ \mathbf{0} = \dfrac{c_2 + x }{2 c + 2 x } A + \dfrac{c_1 + x}{2 c + 2 x} B + \dfrac{c }{2 c + 2 x} C $
where $ c = c_1 + c_2 $
Multiplying through by $(2 c + 2 x ) $ gives
$ \mathbf{0} = (c_2 + x) A + (c_1 + x) B + c C $
Substituting $A = n + c_1 u $ and $ B = n - c_2 u $ gives
$ \mathbf{0} = (c_2 + x) (n + c_1 u) + (c_1 + x) (n - c_2 u) + c C $
which simplifies to
$ \mathbf{0} = (c + 2 x ) n + x (c_1 - c_2 ) u + c C $
Since, $n$ and $u$ are orthogonal unit vectors, it is convenient to find the components of $C$ onto them.
Dot the above expression with $n$. This gives
$ 0 = (c + 2 x) + c C_n $
Similarly, if we dot the above expression with $u$, we'll get
$ 0 = x (c_1 - c_2) + c C_u $
So that
$ C_n = - \dfrac{c + 2 x}{c} $
$ C_u = - \dfrac{x (c_1 - c_2) }{c} $
which are in terms of $x$. To find $x$, note that $|CA| = c_1 + x $, but
$ C = C_n n + C_u u = \left(- \dfrac{c + 2 x}{c}\right) n + \left( - \dfrac{x (c_1 - c_2) }{c} \right) u $
And
$ A = n + c_1 u $
Therefore,
$ C - A = \left( -2 - 2 \dfrac{x}{c} \right) n + \left( \dfrac{x}{c} (c_2 - c_1) - c_1 \right) u $
Hence,
$ (c_1 + x)^2 = \left( -2 - 2 \dfrac{x}{c} \right)^2 + \left( \dfrac{x}{c} (c_2 - c_1) - c_1 \right)^2 $
Solving this quadratic equation gives,
$ \dfrac{x}{c} = \dfrac{1}{c_1 c_2 - 1} $
Therefore, the third vertex coordinates are
$ C = \left(-1 - 2 \dfrac{x}{c} \right) n + \left( \dfrac{x}{c} (c_2 - c_1) \right) u $
Substituting for $ \dfrac{x}{c} $ gives
$ C = \dfrac{ -1 - c_1 c_1 }{c_1 c_2 - 1 } n + \dfrac{ c_2 - c_1 }{ c_1 c_2 - 1} u $
Substituting the followign quantities as derirved above
$ c_1 = \dfrac{1}{\cos \theta} ( \dfrac{1}{e} + \sin \theta ) $
$ c_2 = \dfrac{1}{\cos \theta} ( \dfrac{1}{e} - \sin \theta ) $
$ n = ( - \sin \theta, \cos \theta ) $
$ u = (\cos \theta, \sin \theta ) $
And simplifying gives,
$ C = \left( \sin \theta , - \dfrac{(e^2 + 1)}{ (1 - e^2)} \cos \theta \right) $
This is the equation of an ellipse with semi-minor axis length $1$ (along the $x$ axis) and semi-major axis length of $\dfrac{ (e^2+1)}{(1 - e^2)} $ along the $y$ axis.
Since we started with an ellipse, we have to scale this locus of $C$ (the ellipse with the specified semi-minor and semi-major axes lenths) horizontally and vertically by $a$ and $b$ respectively. This will turn it into an ellipse with a semi-minor (horizontal) axis of length $a$, and a semi-major (vertical) axis of length $b \dfrac{ (e^2+1)}{(1 - e^2)} $.
The locus for the third vertex of the circumscribed triangle about an ellipse with $a = 5, b = 3$ (shown in blue) is the ellipse shown in red in the figure below. The two directrices are shown in black.