# Equal angles formed by the tangent lines to an ellipse and the lines through the foci.

Given an ellipse with foci $F_1, F_2$ and a point $P$. Let $T_1, T_2$ the points of tangency on the ellipse determined by the tangent lines through $P$. Show that $\widehat {T_1 P F_1} = \widehat {T_2 P F_2}$.

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–  Sgernesto Mar 7 '13 at 16:36

let ellipse is: $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, set $P(a,t)$, the tangent line $PT1$ is $y=k(x-a)+t, k=tanA_1$,since it is tangent line, so put in ellipse equation, we have:

$b^2x^2+a^2(kx-ka+t)^2=a^2b^2$, ie,$b^2+a^2k^2)x^2+2a^2k(t-ka)x+a^2(t-ka)^2-a^2b^2=0$

$\Delta$=$4a^4*k^2*(t-ka)^2-4(b^2+a^2*k^2)[a^2*(t-ka)^2-a^2*b^2]$ ,$\Delta=0$,we get $2kta-t^2+b^2=0$, that is, $k=\dfrac{t^2-b^2}{2ta}$

let $k_1$ is$PF1$'s slope,$k_1=\dfrac{t}{a+c}=tanA_2$, the$\angle T1PF1=|A_1-A_2|$, now we caculate $tan(A_1-A_2)$:

$tan(A_1-A_2)$=$\dfrac{tanA_1-tanA_2}{1+tanA_1tanA_2}=\dfrac{k-k_1}{1+k*k_1}=\dfrac{\dfrac{t^2-b^2}{2ta}-\dfrac{t}{a+c}}{1+\dfrac{t^2-b^2}{2ta}\dfrac{t}{a+c}}=\dfrac{(t^2-b^2)(a+c)-2at^2}{2at^2+2act+t^3-tb^2}$

since $b^2=a^2-c^2$,we get :

RHS$=\dfrac{(t^2-a^2+c^2)(a+c)-2at^2}{t(2a^2+2ac+t^2-a^2+c^2)}=\dfrac{c-a}{t}$

let $k_2$ is line $PF2$'s slope, then $k_2=\dfrac{t}{a-c}=tan\angle PF2T2$, since $\angle T2PF2=\dfrac{\pi}{2}-\angle PF2T2$, so $tan \angle T2PF2=\dfrac{1}{tan\angle PF2T2}=\dfrac{a-c}{t}=tan(A_2-A_1)$ , since the angels are acute angle, we have:

$\angle T2PF2=A_2-A_1=\angle T1PF1$

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