MHB Difference between distance from two fixed points is a positive constant

[Dustinsfl]

The set of all points \(P(x,y)\) in a plane, such that the difference of their distance from two fixed points is a positive constant is called?

ellipse
hyperbola
parabola
circle

How do I work this out? Are the two fixed points supposed to be the foci? Wouldn't this also depend on the how one performs the difference? For instance, take a circle of radius a. Then \(-a - a = -2a\) which is negative.

 

[HallsofIvy]

We can simplify by setting up an appropriate coordinate system. We are given two points to be the foci so set up a coordinate system with the x-axis through those points. We are also free to set the y-axis half way between the two foci so that they are at (-a, 0) and (a, 0).

Let (x, y) be a point on the graph. What is the distance from (x, y) to (-a, 0)? What is the distance from (x, y) to (a, 0)? Since you are told that "the difference of their distance from two fixed points is a positive constant" subtract those and set them equal to some constant, R, say. Now simplify, by getting rid of the square roots, until you can identify the type of equation.

 

[Prove It]

Ellipse - Wikipedia, the free encyclopedia

 

[Ackbach]

I think it's an hyperbola.

Circle: locus of points equidistant from a single point (center).

Ellipse: locus of points such that the sum of the distances from the two foci is a constant.

Parabola: locus of points equidistant from a point (vertex) and a line (directrix).

Hyperbola: locus of points such that the difference of the distances from the two foci is a constant.

 

[Dustinsfl]

I think it's an hyperbola.

Circle: locus of points equidistant from a single point (center).

Ellipse: locus of points such that the sum of the distances from the two foci is a constant.

Parabola: locus of points equidistant from a point (vertex) and a line (directrix).

Hyperbola: locus of points such that the difference of the distances from the two foci is a constant.

HallsofIvy never specified a conic just a method to find the conic. Do you agree with that approach? I only ask because when I went through it, I did obtain an ellipse.

 

[Ackbach]

HallsofIvy never specified a conic just a method to find the conic. Do you agree with that approach? I only ask because when I went through it, I did obtain an ellipse.

HoI's approach should work just fine - I think the result should be an hyperbola. Can you show your working?

 

[Dustinsfl]

HoI's approach should work just fine - I think the result should be an hyperbola. Can you show your working?

\begin{align}
\sqrt{(x+a)^2 + y^2} - \sqrt{(x-a)^2 + y^2} &= R\\
\sqrt{(x+a)^2 + y^2} &= R + \sqrt{(x-a)^2 + y^2}\\
x^2 + 2ax + a^2 + y^2 &= R^2 + 2R\sqrt{(x-a)^2 + y^2} + x^2 - 2ax + a^2 + y^2\\
ax - \left(\frac{R}{2}\right)^2 &= \frac{R}{2}\sqrt{(x-a)^2 + y^2}\\
a^2x^2 - 2ax\left(\frac{R}{2}\right) + \left(\frac{R}{2}\right)^4 &= \left(\frac{R}{2}\right)^2x^2 - 2ax\left(\frac{R}{2}\right)^2 + a^2\left(\frac{R}{2}\right)^2 + y^2\left(\frac{R}{2}\right)^2\\
\left(\frac{R}{2}\right)^2 \left(\left(\frac{R}{2}\right)^2 - a^2\right) &= \left(\left(\frac{R}{2}\right)^2 - a^2\right)x^2 + \left(\frac{R}{2}\right)^2y^2
\end{align}
By Pythagoras, \(a^2 + v^2 = z^2\). Let \(z = \frac{R}{2}\).
Then
\[
z^2v^2 = v^2x^2 + z^2y^2\Rightarrow 1 = \frac{x^2}{z^2} + \frac{y^2}{v^2}.
\]

 

[Ackbach]

\begin{align}
\sqrt{(x+a)^2 + y^2} - \sqrt{(x-a)^2 + y^2} &= R\\
\sqrt{(x+a)^2 + y^2} &= R + \sqrt{(x-a)^2 + y^2}\\
x^2 + 2ax + a^2 + y^2 &= R^2 + 2R\sqrt{(x-a)^2 + y^2} + x^2 - 2ax + a^2 + y^2\\
ax - \left(\frac{R}{2}\right)^2 &= \frac{R}{2}\sqrt{(x-a)^2 + y^2}\\
a^2x^2 - 2ax\left(\frac{R}{2}\right) + \left(\frac{R}{2}\right)^4 &= \left(\frac{R}{2}\right)^2x^2 - 2ax\left(\frac{R}{2}\right)^2 + a^2\left(\frac{R}{2}\right)^2 + y^2\left(\frac{R}{2}\right)^2\\
\left(\frac{R}{2}\right)^2 \left(\left(\frac{R}{2}\right)^2 - a^2\right) &= \left(\left(\frac{R}{2}\right)^2 - a^2\right)x^2 + \left(\frac{R}{2}\right)^2y^2
\end{align}

I'm good up to here. I'm not at all sure, though, that $R/2-a>0$. The distance between the two foci is $2a$. Take a point on the line between the two foci that satisfies the requirement. Suppose the distance from one focus is $g$, and the distance from the other focus is $h$, with $h>g$. Then $h-g=R$, as we know the definition of your shape to be. But it is also true that $h+g=2a$, since the point we've picked is on the line between the two foci. That is,
\begin{align*}
h-g&=R \\
h+g&=2a.
\end{align*}
Subtracting the second equation from the first yields $R-2a=-2g<0$, since $g>0$. It follows that $R/2<a$, and hence $R^2/4<a^2$. Therefore,
$$\left(\frac{R}{2}\right)^2 \left(\left(\frac{R}{2}\right)^2 - a^2\right) = \left(\left(\frac{R}{2}\right)^2 - a^2\right)x^2 + \left(\frac{R}{2}\right)^2y^2 $$
describes an hyperbola.