An equation with graph as a line *segment*: $ \sqrt {x^2 + (y+12)^2} = 13 - \sqrt{(x-5)^2 + y^2} $

Question

The graph of this function is (quite oddly) a line segment. I don't understand why it is so. I mean, is there a way of telling just by looking at this equation that its graph will be a line segment? $$ \sqrt {x^2 + (y+12)^2} = 13 - \sqrt{(x-5)^2 + y^2} $$

The solution given said $PA + PB = 13$ (where $P$ is a variable point and $A, B$ are fixed points obvious from the equation). But this kind of equations are for ellipses, right?

Strangely, if you modify this equation just slightly to this: $$ \sqrt {x^2 + (y+12)^2} = 13 - \sqrt{(y-5)^2 + x^2} $$

then it gives a 'real' ellipse. Probably the eccentricity of the previous one plays a role here? And if so, is there a mess-free way of finding out the eccentricity? Trying to get rid of the radicals is painful for this equation.

Answer 1

Think about the description of an ellipse with foci $F_1$ and $F_2$ as the set of all points $P$ such that $PF_1+PF_2 = C$, for some fixed $C$. What happens if $F_1F_2 = C$?

Answer 2

The square root on the left computes the distance from $(0,-12)$, the square root on the right conmputes the distance from $(5,0)$ and the equation expresses that the sum of these two distances is $13$. Since the distance from $(0,-12)$ to $(5,0)$ happens to be precisely $13$, this equality holds precisely for the points on the line segment between these two points. This can also be viewed as a degenerate ellipse. If you replace $13$ with $13.1$, the line segment "thickens" to a proper ellipse.