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I study maths as a hobby. I am working through a text book and am on the section, points, distances and loci. A question asks:

$A$ is the point $(1,0)$ and $B$ the point $(-1,0)$. Find the set of points $P$ such that $PA + PB = 4$.

If I say let coordinates of $P = (x,y)$ , I can see that

$$PA = \sqrt {(x-1)^2 + y^2},\\ PB=\sqrt{(x+1)^2+y^2},$$

but I cannot put this together to find the locus of $P$. I know from looking at the answer that it is an ellipse but cannot derive it myself.

math-physicist
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Steblo
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    See for example https://math.stackexchange.com/q/878091/42969 – Martin R Jul 11 '22 at 15:20
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    That's the definition of an ellipse. – John Douma Jul 11 '22 at 15:25
  • When you say (...)I know from looking at the answer that it is an ellipse but cannot derive it myself.(...), what definition of an ellipse do you have? – mathcounterexamples.net Jul 11 '22 at 15:29
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    I suppose OP wants to derive something like $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. To get such results one needs to square equation $4-\sqrt{(x-1)^2+y^2}=\sqrt{(x+1)^2+y^2}$ to get $8\sqrt{(x-1)^2+y^2}=16-4x$, then divide by 4 and square again to get $3x^2+4y^2=12$ which can be written as $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a=2$, $b=\sqrt{3}$. – Ivan Kaznacheyeu Jul 11 '22 at 16:53
  • $a$ and $b$ can also be obtained from $AB$ and $PA+PB$ like $a=\frac{PA+PB}{2}$ and $b=\frac{\sqrt{(PA+PB)^2-AB^2}}{2}$. – Ivan Kaznacheyeu Jul 11 '22 at 16:56

2 Answers2

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from the definition of an ellipse, the points $(1,0)$ and $(-1,0)$ are precisely the foci (its the two poins for which the sum of distances is constant).

So your focal distance is 2 and your center is $(0,0)$, which means $2c = 2 \implies c = 1$ ($c$ here represents the distance from each focus focus to the center).

Also, since the sum of distances is $4$, we have that $2a=4 \implies a=2$, where $a$ is the length of the major axis (which is in the x-axis in this case, since both of the foci are)

from the equation $a^2=b^2+c^2$ (where $b$ is the minor axis, which will be on top of the $y$ axis), we must have that $b^2 = a^2-c^2 = 2^2-1^2 = 3$. So then we can set up the equation of this ellipse as

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \iff \frac{x^2}{4} + \frac{y^2}{3} =1$

Victor Afonso
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The answer by Victor Afonso assumes that you know the answer is work from ellipse and work backwards. Here, I give an answer that continues your progress.

Indeed, you have all the tools already. The equation you have is

$$ \sqrt{(x - 1)^2 + y^2} + \sqrt{(x + 1)^2 + y^2} = 4 $$

Now, if we directly square both sides, we get some really mess terms on the left under the square root. Instead, we move one term to another side and square it:

$$\begin{align*} \sqrt{(x - 1)^2 + y^2} &= 4 - \sqrt{(x + 1)^2 + y^2} \\ (x - 1)^2 + y^2 &= 16 + (x + 1)^2 + y^2 - 8\sqrt{(x + 1)^2 + y^2} \\ 8\sqrt{(x + 1)^2 + y^2} &= 16 + (x^2 + 2x + 1) - (x^2 - 2x + 1) \\ 2\sqrt{(x + 1)^2 + y^2} &= x + 4 \\ 4(x + 1)^2 + 4y^2 &= x^2 + 8x + 16 \\ &\vdots \\ 3x^2 + 4y^2 - 12 &= 0 \\ \frac{x^2}{4} + \frac{y^2}{3} &= 1 \end{align*}$$

Indeed, this is an ellipse, as others have suggested. I think the nudge you're missing is moving one square root to the other side to avoid having $\fbox{}^4$ terms :) Have fun!

Gareth Ma
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