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Let $\frac{x^2}{a^2}+\frac{y^2}{b^2} =1$ be an ellipse. Let $A = (x_0,y_0)$ be a point outside of an ellipse. Draw two lines that touch an ellipse and denote two tangent points by $D_1 = (x_1,y_1),D_2 = (x_2,y_2)$. My question is that if I draw a line passing through $A$ and the middle point of $D_1$ and $D_2$ i.e., $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$ then it passes the origin.

enter image description here

Here's the image I found on google. I think it's correct but I don't know how to prove this. I already know the equation of a line passing $D_1$ and $D_2$ is $\frac{xx_0}{a^2}+\frac{yy_0}{b^2} =1$. So what I need to show is $$\frac{y_0}{x_0}\left(\frac{x_1+x_2}{2}\right) = \frac{y_1+y_2}{2}$$ But I don't know how to get further. How can I do this? Is there any simple geometric proof of this?

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You can also scale the ellipse either horizontally or vertically so that it becomes a circle, then the image of segment $D_1D_2$ becomes a chord in this circle, so clearly the line segment connecting the image of $A$ to the origin passes through the midpoint of this chord. And since ratios of distances are preserved when scaling, it follows that in the original figure (the ellipse and its tangents), the line connecting $A$ to the origin passes through the center of $D_1 D_2$.

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  • $\begingroup$ That's interesting. But I don't think 'the ratios of distances are preserved when scaling' is a trivial statement. May I ask some proof of that statement? $\endgroup$
    – love_sodam
    Sep 15 '21 at 19:27
  • $\begingroup$ Suppose $R$ is on $AB$, such that $\dfrac{AR}{AB} = t $, then $R = A + t (B - A) $. Apply the scaling, which is linear, it follows that the image of $R$, call it $R'$ is given by: $R' = A' + t (B' - A') $, hence, $\dfrac{ A' R' }{A' B'} = t = \dfrac{AR}{AB} $. $\endgroup$ Sep 15 '21 at 20:43
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ Sep 15 '21 at 21:46
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The equation of the line $D_1 D_2$ which is chord of contact of the ellipse is $$\frac{xx_0}{a^2}+\frac{yy_0}{b^2}=1~~~~~~~(1)$$ $D_1D_2$ is also a chord whose mid point is $(h,k)$, where $h=(x_1+x_2)/2, k=(y_1+y_2)/2$, we have $$\frac{xh}{a^2}+\frac{yk}{b^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}=p~~~~(2)$$ Compare (1) and (2) to get $x_0=ph$ and $y_0=kp$ so we get $$\frac{x_1+x_2}{x_0}=\frac{y_1+y_2}{y_0}$$

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  • $\begingroup$ how did you get the equation (2)? $\endgroup$
    – love_sodam
    Sep 15 '21 at 12:57
  • $\begingroup$ Sorry Eq. (2) was slightly wrong, I have corrected it now. For a conic $S=0$ the equation ofchord whose mid point is $(x;,y')$ is given by $T=S'$ where $T$ is equation of tangent at $(x',y')$. $\endgroup$
    – Z Ahmed
    Sep 15 '21 at 13:44
  • $\begingroup$ Still confusing. What is $S$? $\endgroup$
    – love_sodam
    Sep 15 '21 at 19:28
  • $\begingroup$ Suppose the circle is $S(x,y)=x^2+y^2-a^2=0$ if you find the equation of its chord whose mid point is $(x',y')$ it comes to $xx'+yy'=x'^2+y'^2$ which can be written more generally as $T=S'$ where $T=xx'+yy'-a^2$ and $S'=S(x',y')$. For conics it is generally true. $\endgroup$
    – Z Ahmed
    Sep 16 '21 at 5:05
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Let's start with the equation of the line $D_1D_2$, i.e.

$${xx_0\over a^2}+{yy_0\over b^2}=1$$

$D_1(x_1,y_1)$ and $D_2(x_2,y_2)$ are the intersection points of that line and the ellipse. So they satisfy also the equation of the ellipse.

Substitution and multiplication by $y_0^2/b^2$ gives us that $x_i/a$ are the two solutions of the quadratic

$$X^2\left({y_0^2\over b^2}+{x_0^2\over a^2}\right)-2X{x_0\over a}+1-{y_0^2\over b^2}=0$$

and we deduce from that

$${x_1+x_2\over 2}={x_0\over {x_0^2\over a^2}+{y_0^2\over b^2}}$$

Similarly substitution and multiplication by $x_0^2/a^2$ gives us that $y_i/b$ are the two solutions of the quadratic

$$Y^2\left({x_0^2\over a^2}+{y_0^2\over b^2}\right)-2Y{y_0\over b}+1-{x_0^2\over a^2}=0$$

and thus

$${y_1+y_2\over 2}={y_0\over {y_0^2\over b^2}+{x_0^2\over a^2}}$$

The line passing through the origin and the middlepoint of $D_1D_2$ has the following equation $x_0y-y_0x=0$ and it passes through $M(x_0,y_0)$

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