Given an elipse with two focus $F_{1}$ an $F_{2}$, and $A$ is an arbitrary point at the elipse. Stright line $AF_{1}$ has another intersection point $B$ with the elipse, and $AF_{2}$ has another intersection point $C$ with the elipse.
And point $P$ is the intersection of line $BF_{2}$ and $CF_{1}$, now the problem is to prove that $|PF_{1}|+|PF_{2}|$ is constant.
Thanks in advance!
OK, imagine you have tacks at the points $(-c,0$ and $(c,0)$, which hold each end of a string of length $2 a$. We draw an ellipse by holding a pen taut against the string. The sum of the distances to a point on the ellipse from each of the tack points is
$$\sqrt{(x+c)^2+y^2} + \sqrt{(x-c)^2+y^2} = 2 a$$
The trick is to manage the algebra so that the derivation is readable. First, square both sides to get
$$(x-c)^2 + (x+c)^2 + 2 y^2 + 2 \sqrt{x^2+y^2+c^2+2 c x} \sqrt{x^2+y^2+c^2-2 c x} = 4 a^2$$
This simplifies a little to
$$x^2+y^2+c^2+\sqrt{(x^2+y^2+c^2)^2-4 c^2 x^2} = 2 a^2$$
Now we need to rid ourselves of this remaining square root by isolating it:
$$\begin{align}(x^2+y^2+c^2)^2-4 c^2 x^2 &= [2 a^2 - (x^2+y^2+c^2)]^2\\ &= 4 a^4 - 4 a^2 (x^2+y^2+c^2) + (x^2+y^2+c^2)^2 \end{align}$$
We have some fortuitous cancellation which leaves us with a quadratic. Rearrange to get
$$(a^2-c^2) x^2 + a^2 y^2 = a^2 (a^2-c^2)$$
or, in standard form:
$$\frac{x^2}{a^2} + \frac{y^2}{a^2-c^2} = 1$$
Note that, for an ellipse, $a>c$. We interpret $a$ to be the semimajor axis, $c$ to be the focal length, and $b=\sqrt{a^2-c^2}$ is the semiminor axis.
To prove that the expression for the ellipse has the sum of the distances from the foci being constant, work backward from this sequence.
OK, let's start heavy work as it has a lot of calculations.
let ellipse is $ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 $ ,$c^2=a^2-b^2$, so $F1(c,0),F2(-c,0)$, for any point at ellipse, we can always let $x=acosQ,y=bsinQ$
let $A(acosQ,bsinQ), B(acosQ1，bsinQ1），C(acosQ2,bsinQ2)$
since $A,F1,B$ co-line, so $\dfrac{bsinQ-0}{acosQ-c}=\dfrac{0-bsinQ1}{c-acosQ1}$,ie.$ \dfrac{sinQ}{acosQ-c}=\dfrac{sinQ1}{acosQ1-c} $ ,$asinQcosQ1-csinQ=asinQ1cosQ-csinQ1$,$ a(sinQcosQ1-cosQsinQ1)=c(sinQ-sinQ1)$ , $asin(Q-Q1)=2c*cos\dfrac{Q+Q1}{2}sin\dfrac{Q-Q1}{2} $, $2a*sin\dfrac{Q-Q1}{2}cos\dfrac{Q-Q1}{2}=2c*cos\dfrac{Q+Q1}{2}sin\dfrac{Q-Q1}{2}$,$a*cos\dfrac{Q-Q1}{2}=c*cos\dfrac{Q+Q1}{2}$, $a(cos\dfrac{Q}{2}cos\dfrac{Q1}{2}+sin\dfrac{Q}{2}sin\dfrac{Q1}{2})=c(cos\dfrac{Q}{2}cos\dfrac{Q1}{2}-sin\dfrac{Q}{2}sin\dfrac{Q1}{2}) $,$a(1+tan\dfrac{Q}{2}tan\dfrac{Q1}{2})=c(1-tan\dfrac{Q}{2}tan\dfrac{Q1}{2}),(a+c)tan\dfrac{Q}{2}tan\dfrac{Q1}{2}=c-a$,now we get a important equation : $tan\dfrac{Q1}{2}=\dfrac{c-a}{c+a}cot\dfrac{Q}{2}$
sine $A,F2,C $ co-line,so $\dfrac{bsinQ-0}{acosQ+c}=\dfrac{0-bsinQ2}{-c-acosQ2}$,with similar work, we get $tan\dfrac{Q2}{2}=\dfrac{c+a}{c-a}cot\dfrac{Q}{2}$ .
for line $BF2$, we have $\dfrac{y-0}{x+c}=\dfrac{0-bsinQ1}{-c-acosQ1}$,RHS$ =\dfrac{bsinQ1}{acosQ1+c}=\dfrac{b*\dfrac{2tan\dfrac{Q1}{2}}{1+tan^2\dfrac{Q1}{2}}}{a*\dfrac{1-tan^2\dfrac{Q1}{2}}{1+tan^2\dfrac{Q1}{2}}+c} $=$\dfrac{2b*tan\dfrac{Q1}{2}}{a+c+(c-a)tan^2\dfrac{Q1}{2}}$, now we replace $tan\dfrac{Q1}{2}$,
RHS=$\dfrac{2b*\dfrac{c-a}{c+a}*cot\dfrac{Q}{2}}{c+a+(c-a)(\dfrac{c-a}{c+a})^2*cot^2\dfrac{Q}{2}}$=$\dfrac{2(c^2-a^2)b*cot\dfrac{Q}{2}}{(a+c)^3+(c-a)^3*cot^2\dfrac{Q}{2}}$=$\dfrac{-2b^3*sin\dfrac{Q}{2}cos\dfrac{Q}{2}}{a^3(sin^2\dfrac{Q}{2}-cos^2 \dfrac{Q}{2})+3a^2c+c^3+3ac^2(sin^2\dfrac{Q}{2}-cos^2 \dfrac{Q}{2})}$=$\dfrac{-b^3sinQ}{c(3a^2+c^2)-a(a^2+3c^2)cosQ}$, then
$y=\dfrac{-b^3sinQx}{c(3a^2+c^2)-a(a^2+3c^2)cosQ}+\dfrac{-b^3c*sinQ}{c(3a^2+c^2)-a(a^2+3c^2)cosQ}$------formula 1,
for line CF1: $\dfrac{y-0}{x-c}=\dfrac{0-bsinQ2}{c-acosQ2}$,with same work, we get RHS=$\dfrac{b^3sinQ}{c(c^2+3a^2)+a(a^2+3c^2)cosQ}$, so we have
$y=\dfrac{b^3sinQx}{c(3a^2+c^2)+a(a^2+3c^2)cosQ}-\dfrac{b^3c*sinQ}{c(3a^2+c^2)+a(a^2+3c^2)cosQ}$------formula 2,
with two formulas, we get $ x= \dfrac{-a(a^2+3c^2)cosQ}{3a^2+c^2} ,y=\dfrac{-b^3sinQ}{3a^2+c^2}$,if you want feel easy, let $Q=\pi+Q3$, then $x= \dfrac{a(a^2+3c^2)cosQ3}{3a^2+c^2} ,y=\dfrac{b^3sinQ3}{3a^2+c^2}$,
let $a_1=\dfrac{a(a^2+3c^2)}{3a^2+c^2},b_1=\dfrac{b^3}{3a^2+c^2}$,
we finally get for$ P(x,y), x=a_1*cosQ3,y=b_1*sinQ3 $,which means P is on a ellipse,the focus c1 should be :$c_1^2=a_1^2-b_1^2$,again we have some work to do,
RHS=$[\dfrac{a(a^2+3c^2)}{3a^2+c^2}]^2-[\dfrac{b^3}{3a^2+c^2}]^2$=$\dfrac{[a^3+3ac^2)]^2-b^6}{(3a^2+c^2)^2}=\dfrac{a^6+6a^4c^2+9a^2c^4-(a^2-c^2)^3}{(3a^2+c^2)^2}=\dfrac{c^2(c^4+6a^2c^2+9a^4)}{(3a^2+c^2)^2}=c^2$,
it means it has same focus (c,0),(-c,0). that is all.
Let $AF_1=m,AF_2=n$,$BF_1=p,CF_2=q$.
Lemma$$\frac1m+\frac1p=\frac1n+\frac1q=\frac{2a}{b^2}$$
by def :$$m+n=p+BF_2=q+CF_1=2a$$
Using Menelaus' theorem and $CP=CF_1-F_1P$,we have $$\frac{PF_1\cdot BA\cdot F_2C}{F_1B\cdot AF_2\cdot CP}=1$$ ,we Omit tedious simplification $$PF_1=\frac{\frac{4a^2n}{b^2}-2a-n}{\frac{4a^2}{b^2}-1}$$ Similarly, $$PF_2=\frac{\frac{4a^2m}{b^2}-2a-m}{\frac{4a^2}{b^2}-1}$$ then $$PF_1+PF_2=\frac{2a(4a^2-3b^2)}{4a^2-b^2}$$
