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Given ellipse(${x^2 \over a^2}+{y^2 \over b^2}=1$) and a point $(x_0,y_0)$
We draw two tangent lines to the ellipse that are going through $(x_0,y_0)$ Find the equation of the straight line connecting two points where tangent lines are touching ellipse.
I can do it through solving system of 4 equations. But I think there is an easier, maybe geometric way of solving the problem. Share some new ideas. Thanks.

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Let $(p,q)(s,t)$ be tangent points on the ellipse. Then, the tangent line at $(p,q)$ is $$y-q=-\frac{b^2p}{a^2q}(x-p),$$ i.e. $$a^2qy+b^2px=a^2b^2$$ because $\frac{p^2}{a^2}+\frac{q^2}{b^2}=1,$ i.e. $b^2p^2+a^2q^2=a^2b^2$.

Since this line passes through $(x_0,y_0)$, we have $$a^2qy_0+b^2px_0=a^2b^2\tag1$$ Similarly, we have $$a^2ty_0+b^2sx_0=a^2b^2\tag2$$

From $(1)-(2)$, $$a^2y_0(q-t)+b^2x_0(p-s)=0\quad\Rightarrow \quad \frac{q-t}{p-s}=-\frac{b^2x_0}{a^2y_0}\tag3$$

Also, $$\frac{sq-pt}{s-p}=\frac{s}{s-p}\left(\frac{a^2b^2-b^2px_0}{a^2y_0}\right)-\frac{p}{s-p}\left(\frac{a^2b^2-b^2sx_0}{a^2y_0}\right)=\frac{a^2b^2(s-p)}{a^2y_0(s-p)}=\frac{b^2}{y_0}\tag4$$

From $(3)(4)$, the answer is $$y-q=\frac{q-t}{p-s}(x-p)\iff y=\frac{q-t}{p-s}x+\frac{sq-pt}{s-p}$$ $$\iff \color{red}{y=-\frac{b^2x_0}{a^2y_0}x+\frac{b^2}{y_0}}.$$

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