# How to prove this hypothesis regarding slopes and ellipses?

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.

• What is meant by $k_{QB}$ and $k_{PA}$ ?? May 10, 2022 at 15:12
• The slope of QB and PA, respectively. (That is, $k_{QB}=\dfrac{y_Q-y_B}{x_Q-x_B}$ May 10, 2022 at 15:18
• I'll just consider when M is within it. May 10, 2022 at 15:19
• $x_p=a\cos t_p$, $y_p=b\sin t_p$, $x_q=a\cos t_q$, $y_q=b\sin t_q$, $\frac{0-y_p}{y_q-y_p}=\frac{\frac{a}{\lambda}-x_p}{x_q-x_p}$. Then you can express $k_{PA}$, $k_{QB}$, $\lambda$ in terms of $t_p$ and $t_q$ and check your hypothesis. May 10, 2022 at 16:21
• I have numerically verified your hypothesis, it is correct. May 10, 2022 at 18:22

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

$$\frac{k_{BQ}}{k_{AP}} = \frac{\dfrac{a^2}m+a}{\dfrac{a^2}m-a}=\frac{a+m}{a-m}=\frac{\lambda +1}{\lambda-1}$$