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Let $a, b\in \mathbf{R}^+, \lambda >1$. $\Omega: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, Point $M(\dfrac a{\lambda}, 0), A(-a,0),B(a, 0)$. Let line $l$ pass through $M$ and intersect with $\Omega$ at points $P$ and $Q$. ($P$ is above the $X$-axis while $Q$ is below it). I found that $\dfrac{k_{QB}}{k_{PA}}=\dfrac{\lambda+1}{\lambda-1}$, but cannot prove it.
I tried to prove it in a circle first, but the calculation is just too complex for me to handle.
We can begin by assuming (temporarily) that $b = a$ so that the ellipse is a circle, allowing the use of some convenient circle theorems.
In the figure below I have "stretched" your figure vertically to make a circle and have added some line segments to construct triangles $\triangle APB$ and $\triangle AQB.$
Note that since $AB$ is a diameter of the circle, these two triangles are both right triangles with right angles at $P$ and $Q$ respectively. Therefore $k_{QB} = \tan(\angle ABQ) = AQ/BQ$ and $k_{PA} = \tan(\angle BAP) = BP/AP.$
By the inscribed angle theorem, we can show that triangles $\triangle AMP$ and $\triangle QMB$ are similar (with corresponding vertices listed in those respective orders, for example, the angle at $A$ in $\triangle AMP$ is congruent to the angle at $Q$ in $\triangle QMB$). Let the ratio of lengths in the two triangles be $r$, that is, \begin{align} BQ &= r AP, \\ BM &= r MP, \\ MQ &= r AM. \end{align}
Likewise, we can show that triangles $\triangle BMP$ and $\triangle QMA$ are similar (with corresponding vertices listed in those respective orders). Let the ratio of the triangles be $s$, that is, \begin{align} BP &= s AQ, \\ BM &= s MQ, \\ MP &= s AM. \end{align}
Then $$ \frac{AM}{BM} = \frac{\frac1r MQ}{s MQ} = \frac{1}{rs}. $$ Moreover, \begin{align} \frac{k_{QB}}{k_{PA}} &= \frac{AQ/BQ}{BP/AP} \\ &= \frac{AQ\cdot AP}{BP\cdot BQ} \\ &= \frac{AQ\cdot AP}{(s AQ)\cdot (r AP)} \\ &= \frac{1}{rs} \\ &= \frac{AM}{BM} \\ &= \frac{a + \frac a\lambda}{a - \frac a\lambda} \\ &= \frac{\lambda + 1}{\lambda - 1}. \end{align}
That proves the theorem when $b = a$. For the case when $b \neq a,$ we scale the entire figure vertically by a factor of $b/a.$ This changes the slopes of all lines, but preserves the ratio of the slopes of lines $AP$ and $BQ$, so it is still true that $$ \frac{k_{QB}}{k_{PA}} = \frac{\lambda + 1}{\lambda - 1}. $$
Let's take the coordinates of the points $P$ and $Q$ to be $P(x_P,y_P)$ and $Q(x_Q,y_Q)$ respectively. Since line $l$ passes through the points $P$, $M$ and $Q$, one can write (eq. (I)): $$\frac{y_P}{x_P-\frac{a}{\lambda}}=\frac{y_Q}{x_Q-\frac{a}{\lambda}}$$ and, $$\frac{y_Q}{y_P}=\frac{\lambda x_Q-a}{\lambda x_P-a}$$ For the slopes of the lines $QB$ and $PA$, we have, $k_{QB}=\frac{y_Q}{x_Q-a}$ and $k_{PA}=\frac{y_P}{x_P+a}$. Let's take $R$ to be: $$R=\frac{k_{QB}}{k_{PA}}=\frac{y_Q(a+x_P)}{y_P(x_Q-a)}$$ Using eq. (I), $$R=\frac{y_Q(a+x_P)}{y_P(x_Q-a)}=\frac{(\lambda x_Q-a)(x_P+a)}{(\lambda x_P-a)(x_Q-a)}$$ The coordinates of the points $P$ and $Q$ satisfy the equation of the ellipse. Therefore, we have, $$y_Q=-b\sqrt{1-\frac{x_Q^2}{a^2}}$$ The negative sign is there because $Q$ is said to be below the x-axis. Similarly, for point $P$, $$y_P=b\sqrt{1-\frac{x_P^2}{a^2}}$$ Using eq. (I), one gets, $$\frac{y_Q}{y_P}=\frac{-\sqrt{a^2-x_Q^2}}{\sqrt{a^2-x_P^2}}=\frac{\lambda x_Q-a}{\lambda x_P-a}$$ and, $$\frac{a^2-x_Q^2}{a^2-x_P^2}=\frac{(\lambda x_Q-a)^2}{(\lambda x_P-a)^2}$$ By calculating this, we come to, $$a\lambda^2(x_P^2-x_Q^2)-2a^2\lambda (x_P-x_Q)+a(x_P^2-x_Q^2)-2\lambda x_Px_Q(x_P-x_Q)=0$$ We can factor $(x_P-x_Q)$ and we will have, $$(x_P-x_Q)[a\lambda^2(x_P+x_Q)-2a^2\lambda +a(x_P+x_Q)-2\lambda x_Px_Q]=0$$ The case where $$x_P-x_Q=0$$ and so $l$ is vertical is easy to prove. Let's analyze the case where the terms in brackets are zero. $$a\lambda^2(x_P+x_Q)-2a^2\lambda +a(x_P+x_Q)-2\lambda x_Px_Q=0$$ By expanding and rewriting the equation, $$a\lambda^2x_Q-a^2\lambda-\lambda x_Px_Q+ax_P=-a\lambda^2 x_P+a^2\lambda+\lambda x_Px_Q-ax_Q$$ Now, we add the terms $\lambda^2x_Px_Q$, $-a\lambda x_P$, $-a\lambda x_Q$ and $a^2$ to both sides of the equation: $$a\lambda^2x_Q+\lambda^2x_Px_Q-a^2\lambda-a\lambda x_P-a\lambda x_Q+a^2-\lambda x_Px_Q+ax_P=-a\lambda^2 x_P+\lambda^2x_Px_Q+a^2\lambda-a\lambda x_P-a\lambda x_Q+a^2+\lambda x_Px_Q-ax_Q$$ By factoring $(\lambda -1)$ in the left and $(\lambda +1)$ in the right sides of the equation, we get, $$(\lambda -1)(a\lambda x_Q+\lambda x_Qx_P-a^2-ax_P)=(\lambda+ 1)(\lambda x_Px_Q-a\lambda x_P-ax_Q+a^2)$$ and then,$$\frac{a\lambda x_Q+\lambda x_Qx_P-a^2-ax_P}{\lambda x_Px_Q-a\lambda x_P-ax_Q+a^2}=\frac{\lambda+1}{\lambda-1}$$ The left side of the equation can be written as, $$\frac{(\lambda x_Q-a)(x_P+a)}{(\lambda x_P-a)(x_Q-a)}=\frac{\lambda+1}{\lambda-1}$$ According to eq. (I), the left side is $R$. Therefore, one can write, $$R=\frac{k_{QB}}{k_{PA}}=\frac{\lambda+1}{\lambda-1}$$
This solution uses the (somewhat advanced) concept of poles and polars.
Assume $M(m,0)$ is the pole, then the equation of the polar $ST$ is
$$ \frac{xm}{a^2}=1 $$
Let $ST$ intersect the X axis at $K$, we have
$$ k_{AP}=\tan \angle SAK \\ k_{BQ}=\tan \angle SBK $$
Thus
$$ \frac{k_{BQ}}{k_{AP}} = \frac{\dfrac{a^2}m+a}{\dfrac{a^2}m-a}=\frac{a+m}{a-m}=\frac{\lambda +1}{\lambda-1} $$
