# [SOLVED]difference between distance from two fixed points is a positive constant

#### dwsmith

##### Well-known member
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

##### Well-known member
MHB Math Helper
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.

MHB Math Helper

#### Ackbach

##### Indicium Physicus
Staff member
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.

#### dwsmith

##### Well-known member
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

##### Indicium Physicus
Staff member
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?

#### dwsmith

##### Well-known member
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
$z^2v^2 = v^2x^2 + z^2y^2\Rightarrow 1 = \frac{x^2}{z^2} + \frac{y^2}{v^2}.$

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#### Ackbach

##### Indicium Physicus
Staff member
\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.