Axes-intersections of normal tangents to an ellipse 
*

*Question: What values can $x_T$,$y_T$,$x_N$, and $y_N$ take on?
Let $T$ and $N$ be the tangent and normal lines to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ at any point on the ellipse in the first quadrant. 
Let $x_T$ and $y_T$ be the $x-$ and $y$-intercepts of $T$ and $x_N$ and $y_N$ be the intercepts of $N$
As $P$ moves along the ellipse in the first quadrant(but not on the axes), what values can $x_T$,$y_T$,$x_N$, and $y_N$ take on?



*

*What I've managed thus far
Here is a graph of one of the scenarios, 



*

*Slopes of lines T and N: 


*

*Slope of $T$ is equal to $y\prime = -\frac{4x}{9y}$

*Slope of $N$ is equal to $-\frac{1}{y\prime} = \frac{9y}{4x}$


*Line equations for lines $T$ and $N$:


*

*Line $T:\quad y - y_T = -\frac{4x}{9y}\Big(x-x_T\Big) \Leftrightarrow y = -\frac{4x^2}{9y} + \frac{4xx_T}{9y} + y_T$

*Line $N:\quad y - y_N = \frac{9y}{4x}\Big(x - x_N\Big) \Leftrightarrow y = \frac{9y}{4} - \frac{9yx_N}{4x} + y_N$


*Now I will take a few limits on the line equations of $T$ and $N$:

Limits on $T$


*

*$\lim_{x \to 0} (y) = \lim_{x \to 0}\Big(-\frac{4x^2}{9y} + \frac{4xx_T}{9y} + y_T\Big)$
That gives me $$\lim_{x\to 0} y_T = 2$$

*Multiplying line $T$'s equation by $9y$, I obtain the following:
$$\lim_{x\to 3} x_T = 3$$

Limits on $N$


*

*$\lim_{x \to 3} (y)  = \lim_{x \to 3} \Big(\frac{9y}{4} - \frac{9yx_N}{4x} + y_N\Big)$


*

*That gives me, $$\lim_{x \to 3} y_N = 0$$


*Multiplying line $N$'s equation by $4x$, I obtain the following: $$\lim_{x\to 0} x_N = 0$$

That's where I'm stuck, I still need to find $\lim_{x\to 0} x_T$, $\lim_{x\to 0} y_N$, $\lim_{x\to 3} y_T$, and $\lim_{x\to 3} x_N$.
That is, at the moment I have partial ranges for $x_T$, $y_T$, $x_N$, and$y_N$, as follows $$x_T(3, \lim_{x\to 0} x_T)$$
$$y_T(2, \lim_{x\to 3} y_T)$$
$$x_N(0, \lim_{x\to 3} x_N)$$
$$y_N(0, \lim_{x\to 0} y_N)$$

Does anybody have any hints, suggestions, or alternative approaches?
 A: Let us denote $P=(x,y)$ and $(X,Y)$ the variables in $\mathbb R^2$. Then the tangent line is
$$
T:\quad Y-y=-\frac{4x}{9y}(X-x)\quad\text{or}\quad 4xX+9yY=4x^2+9y^2=36
$$
(because $(x,y)$ verifies the equation of the ellipse). The points lying in $T$ are $(x_T,0)$ and $(0,y_T)$ and substituting $X=x_T,Y=0$ and $X=0,Y=y_T$ we get
$$
4xx_T=36\quad\text{and}\quad 9yy_T=36,
$$
that is:
$$
x_T=\frac{9}{x}\quad\text{and}\quad y_T=\frac{4}{y}.
$$
From this one gets the limits (that can be $\infty$). The geometric meaning of $\infty$ is that the tangent line is parallel to the corresponding axis.
For the normal one starts with the true equation:
$$
N: \quad Y-y=\frac{9y}{4x}(X-x)\quad\text{or}\quad 9yX-4xY=5xy.
$$
Since $(x_N,0)$ and $(0,y_N)$ belong to $N$ we have, respectively
$$
9yx_N=5xy\quad\text{and}\quad -4xy_N=5xy,
$$
that is
$$
x_N=\frac{5x}{9}\quad\text{and}\quad y_N=-\frac{5y}{4},
$$
from which one computes the limits. In this case the limit normals coincide with the $X$-axis for $x\to3$ or the $Y$-axis for $x\to 0$. The limits are
$$
\begin{cases}\lim_{x\to3}x_N=\frac{5}{3},\\ \lim_{x\to3}y_N=0,\end{cases}\quad\text{and}\quad
\begin{cases}\lim_{x\to0}x_N=0,\\ \lim_{x\to0}y_N=-\frac{5}{2},\end{cases}
$$ 
and the geometric meaning is not that clear. But one can say this. First, the distance between the points 
$(x_N,0)$ and $(0,y_N)$ is (using again $4x^2+9y^2=36$):
$$
d=\sqrt{x_N^2+y_N^2}=\sqrt{\frac{25x^2}{81}+\frac{25y^2}{16}}=
\frac{5}{18}\sqrt{81-5x^2},
$$
so that 
$$
\lim_{x\to3}d=\frac{5}{3}\quad\text{and}\quad\lim_{x\to0}d=\frac{5}{2}
$$
Thus if one accepts the value cero of the limits above as geometrically natural, the other limits follow.
Another meaning of this maybe surprising limits is that the normals to the ellipse (in the first quadrant) do not cut the $X$-axis further than some point in that axis (namely, $\frac{5}{3}$) and do not cut the $Y$-axis lower than another (namely, $-\frac{5}{2}$).
