Spurious roots - graphical reason for invalidity

Question

A question goes;

Two parabolae have the same focus. If that their directrices are the x-axis and y-axis respectively, then find the slope of their common chord.

To solve it, you write out the general equations $$(x-a)^2=4(\frac{1}{2}b)(y-\frac{b}{2}) \\(y-b)^2=4(\frac{1}{2}a)(x-\frac{a}{2})$$ Where a and b are constants.

Subtracting these give you $x^2-y^2=0$, or $(x+y)(x-y)=0$.

It can be shown that at one time, two parabolae satisfying these properties can only intersect at two points. So the slope of the common chord is either 1 or -1, as obtained just above.

The problem is, that seems to be a result that should depend on the signs of a and b. I played around with this applet for a while, and that's how it plays out.

If I had a specific a and b, say 3 and 5, the solution would have one spurious root; $x+y=0$. But I can't work out a reason for why I would get that root too when I carry out the same steps with these two specific parabolae. Similarly, negative values of a and b would mean $x-y=0$ is a spurious root.

How exactly do these invalid roots end up in this analysis? A correct, rigorous treatment would work out so that the result you obtain specifies each root to be valid in specific intervals of a and b, as they are. Where did I mess up in my attempt here?

Answer 1

Is the answer not slope=1? There should be infinite answers for x and y at intersection points since there are an infinite number of equations that satisfy the conditions. A slope of 1 intuitively makes sense as well considering the question is forcing symmetric vertices across the y=x (or y=-x) line along with perpendicular directrices.

Edit: The slope could very well be -1; depending on which quadrant the focus lies in.

Answer 2

The number of constants in the equations can be reduced a little bit:

\begin{align} (x-a)^2 &= b(2y - b), \\ (y-b)^2 &= a(2x - a). \end{align}

We can also write the equations:

\begin{align} (x-a)^2 + b^2 &= 2by, \\ (y-b)^2 + a^2 &= 2ax. \end{align}

Assuming $a\neq 0$ and $b\neq 0,$ the left hand sides of the last two equations are positive, and therefore $x$ has the same sign as $a$ and $y$ has the same sign as $b.$

If $a$ and $b$ have the same sign, then so do $x$ and $y$ and it is not possible that $x + y = 0.$ Therefore the equation $x^2 - y^2 = 0$ is solved only when $x - y = 0.$

If $a$ and $b$ have opposite signs, then so do $x$ and $y$ and it is not possible that $x - y = 0.$ Therefore the equation $x^2 - y^2 = 0$ is solved only when $x + y = 0.$

Graphically, since $x$ must have the same sign as $a$ and $y$ must have the same sign as $b,$ any point $(x,y)$ that is at an intersection of the parabolas must be in the same quadrant as $(a,b).$