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Let $a$ be the largest x-value of an ellipse centered at the origin. Similarly, let $b$ be the largest y-value of the ellipse. Assume that $a>b$. We know that the foci points will be at $c_1=(-\sqrt {a^2-b^2},0)$ and $c_2=(+\sqrt {a^2-b^2},0)$ respectively. Then select some point on the ellipse $p_1$ with x-value $x_1$ between $-a$ and $a$. If we only consider the positive y-values of the ellipse, then we know from the equation of an ellipse that the y-value of $p_1$, will be $y_1=\sqrt{b^2(1-\frac{x_1^2}{a^2}})$ or better yet $y_1=\frac{b\sqrt{a^2-x_1^2}}{a}$. So for a generic x-value, the point on the ellipse will be $p_1=\left(x_1, \frac{b\sqrt{a^2-x_1^2}}{a}\right)$.

I'd like to show that the sum distance of $p_1$ to $c_1$ and $p_1$ to $c_2$ will be a constant that does not depend on the value of $x_1$. To do so, I planned to use Euclid's Metric, and have cancellation of the $x_1$ terms algebraically.

In other words, I would like to show a formula in terms of only $a$ and $b$ for the distance of a point on the ellipse to each foci. I am looking for how to manipulate Euclid's metric to show this cancellation.

For thoroughness, I'll give the original equation for the generic distance from $p_1$ to $c_1$: $\sqrt{(x_1-\sqrt{a^2-b^2})^2+\left(\frac{b\sqrt{a^2-x_1^2}}{a}\right)^2}$.

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  • $\begingroup$ Set $s(x_1)=\sqrt{(x_1-\sqrt{a^2-b^2})^2+\left(\frac{b\sqrt{a^2-x_1^2}}{a}\right)^2}+ \sqrt{(x_1+\sqrt{a^2-b^2})^2+\left(\frac{b\sqrt{a^2-x_1^2}}{a}\right)^2}$ and square both sides. A few of algebra will lead to $s(x_1)=2a.$ $\endgroup$
    – user376343
    Nov 29, 2018 at 23:19
  • $\begingroup$ I am not seeing the steps. Could you show them? $\endgroup$
    – EDS
    Nov 30, 2018 at 0:19
  • $\begingroup$ Is the ellipse also axis-aligned? Centered at the origin doesn’t guarantee this. $\endgroup$
    – amd
    Nov 30, 2018 at 0:48
  • $\begingroup$ @amd3 , this follows from the y-coordinate of the point $p_1$ o the ellipse. $\endgroup$
    – user376343
    Nov 30, 2018 at 3:57

2 Answers 2

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Every point of elipse $(bx)^2 +(ay)^2 =(ab)^2 $ can be represent as $P=(a\cos t , b\sin t)$ hence $$c_1 P +c_2 P =\sqrt{(\sqrt{a^2 -b^2} +a\cos t )^2 + b^2\sin^2 t } +\sqrt{(\sqrt{a^2 -b^2} -a\cos t )^2 + b^2\sin^2 t }=\sqrt{(a+\sqrt{a^2 -b^2}\cos t)^2 } +\sqrt{(a-\sqrt{a^2 -b^2}\cos t)^2 } =2a$$

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  • $\begingroup$ I definitely agree that using the parametric recipe for an ellipse makes this demonstration a lot easier! $\endgroup$ Nov 30, 2018 at 18:13
  • $\begingroup$ Could you show how you got from the first radical to the 2nd? How does the b^2sin^2 cancel out? $\endgroup$
    – EDS
    Dec 12, 2018 at 14:20
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An ellipse is defined as having the property you are trying to prove. The common equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is just an equation that satisfies this property. To show that this is the case, we can use the formula you mentioned: $c=\sqrt{a^2-b^2}$ which is a little bit cleaner as $b^2=a^2-c^2$. We can multiply both sides of the equation of the ellipse by $a^2b^2$ to get $b^2x^2+a^2y^2=a^2b^2$ and then substituting for $b^2$ we get $(a^2-c^2)x^2+a^2y^2=a^2(a^2-c^2)$ and after distributing: $a^2x^2-c^2x^2+a^2y^2=a^4-a^2c^2$. If we add $-2a^2cx$ from both sides we obtain $a^2x^2-c^2x^2+a^2y^2-2a^2cx=a^4-a^2c^2-2a^2cx$. Reordering terms a little bit yields $a^2x^2-2a^2cx+a^2c^2+a^2y^2=a^4-2a^2cx+c^2x^2$. We can factor $a^2$ out of the left side to get $a^2(x^2-2cx+c^2+y^2)=a^4-2a^2cx+c^2x^2$ and we can factor the perfect square in the parentheses and the perfect square on the right to get $a^2((x-c)^2+y^2)=(a^2-cx)^2$. Now we can take a square root on both sides to get $a\sqrt{(x-c)^2+y^2}=a^2-cx$.Multiplying by $-4$ gives $-4a\sqrt{(x-c)^2+y^2}=4cx-4a^2$. Adding $x^2+y^2+c^2-2cx+4a^2$ to both sides gives $4a^2-4a\sqrt{(x-c)^2+y^2}+x^2-2cx+c^2+y^2=x^2+2cx+c^2+y^2$. We can factor the perfect squares on the right and left to get $4a^2-4a\sqrt{(x-c)^2+y^2}+(x-c)^2+y^2=(x+c)^2+y^2$. If you look carefully at the left side you might notice it is a perfect square. Factoring it yields $(2a-\sqrt{(x-c)^2+y^2})^2=(x+c)^2+y^2$. We can take square roots of both sides to get $2a-\sqrt{(x-c)^2+y^2}=\sqrt{(x+c)^2+y^2}$. Reordering terms gives $\sqrt{(x-c)^2+y^2}+\sqrt{(x+c)^2+y^2}=2a$. The first square root is the distance from any point $(x,y)$ to the focus on the right, and the second root is the distance from the same point to the focus on the left. We see that these distances sum to the constant $2a$.

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  • $\begingroup$ Excellent, exactly what I was looking for $\endgroup$
    – EDS
    Dec 1, 2018 at 18:14

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