Finding the locus of the third vertex of the triangle under the given conditions

Question

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?

Answer 1

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 ) $

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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.