Coordinate Geometry (Parabola)

Question

The standard forms of a parabola are normally Y^2=4ax and X^2=4ay. Now if we consider the graphs of these curve, the vertex of each one of them is the intersection of the axes y=0;x=0. Coincidentally they(y=0, x=0) are also found in the equation as. (Y-0)^2=4A(X-0), where they (y=0, x=0)represent lines. If we transpose the term 4A(X-0) to the left we get (Y-0)^2-4A(X-0)=0, this may represent a function pair of straight lines — function of a line, the whole equated to zero . Now analogously is it correct to say (y-0)(y-15)=4a(x-12) as a parabola? Also what could that mean for Locus of the parabola and what are its limitations?

Answer 1

A locus is a set of points. In the context of polynomial curves, a zero locus of a polynomial function $f$ is the set of points $(x,y)$ at which the equation $f(x,y)=0$ is true.

But the zero locus of a function is still a set of points; it is not the function itself. While a function has only one zero locus, a curve in the plane is the zero locus of many functions. If $C$ is the zero locus of $f,$ then $C$ is also the zero locus of the function $g$ where $g(x,y)=kf(x,y),$ for an arbitrary non-zero factor $k.$

For example, the zero locus of $f_1(x,y) = x^2 + y^2 - 4$ is a circle of radius $2$ with center at $(0,0).$ But the zero locus of $f_2(x,y) = 10x^2 + 10y^2 - 40$ is the same circle. So is the zero locus of $f_3(x,y) = 4 - x^2 - y^2.$

Once more, a locus is a set of points, not an equation, polynomial, or other function of numbers, so it does not make sense to talk about the "sum" or "difference" of one locus with another, unless you mean "union" or "set difference" (which is obviously not what you want to talk about).

You can talk about the sum or difference of two functions. Then you can ask, given two polynomial functions of two variables, what are the zero loci of these functions in a Cartesian plane, and what are the zero loci of the sum and difference of those functions?

For example, consider the functions $g_1$ and $h_1$ defined by \begin{align} g_1(x,y) &= 3x^2 + 3y^2 - 12, \\ h_1(x,y) &= \left(x-2\right)^2+y^2-4. \end{align} The zero locus of $g_1$ is the same as the zero set of the previously-mentioned $f_1,$ a circle of radius $2$ centered at $(0,0).$ The zero locus of $h_1$ is a circle of radius $2$ centered at $(2,0).$

The sum of $g_1$ and $h_1,$ which we can write as $g_1+h_1,$ is another polynomial function of two variables. Specifically, the function $p_1 = g_1 + h_1$ is defined by $$p_1(x,y) = \left(3x^2 + 3y^2 - 12\right) + \left(\left(x-2\right)^2+y^2-4\right).$$ The zero locus of this function is a circle with center at $\left(\frac12,0\right).$

The difference of $g_1$ and $h_1$ is written $g_1 - h_1.$ If $m_1 = g_1 - h_1$ then $$m_1(x,y) = \left(3x^2 + 3y^2 - 12\right) - \left(\left(x-2\right)^2+y^2-4\right).$$ The zero locus of this function is a circle with center at $\left(-1,0\right).$

So in this case we have two polynomial functions, each of which produces a circle as its zero locus; the sum produces a third circle, and the difference produces a fourth circle, not a straight line.

However, the simple sum and difference are not the only interesting ways to combine two polynomial functions. We can also take any linear combination of the two functions, $ag_1 + bh_1,$ where $a$ and $b$ are constant factors. If $a$ and $b$ are not both zero, the function $ag_1 + bh_1$ will be another quadratic function of two variables.

For example, let's try $a=1,$ $b=-3.$ That is, let $q_1 = g_1 - 3h_1,$ so $$q_1(x,y) = \left(3x^2 + 3y^2 - 12\right) - 3\left(\left(x-2\right)^2+y^2-4\right).$$ The right-hand side of this equation can be simplified to obtain $$ q_1(x,y) = 12(x - 1), $$ whose zero locus is the line $x = 1.$

Looking further into the last few functions, we find that the zero loci of the five functions $g_1,$ $h_1,$ $g_1 + h_1,$ $g_1 - h_1,$ and $g_1 - 3h_1$ all pass through two points, $\left(1,\sqrt3\right)$ and $\left(1,-\sqrt3\right).$ That is not a coincidence; once we have found that the zero loci of $g_1$ and $h_1$ intersect at two points, we know that $g_1(x,y)=0$ and $h_1(x,y)=0$ simultaneously at those two points, and it follows that any linear combination of $g_1$ and $h_1$ (including their sum and difference) will be zero at those two points, hence its zero locus must include those two points.

One thing about the quadratic functions that produce circles, unlike other quadratic functions, is that the coefficients of the $x^2$ and $y^2$ terms in one such function are always equal. So if we take the functions for two circles, there will always be some linear combination that cancels the $x^2$ terms and cancels the $y^2$ terms, and then (provided that the two circles were not identical or concentric in the first place) you have the equation of a line. However, this line is not necessarily obtained by simply taking the difference of the two functions; it might be some other linear combination, even the sum of the two functions.

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That's the story for two circles that intersect in two points. The story for a line that intersects a circle in two points is similar; by taking linear combinations of a function whose zero locus is the line and a function whose zero locus is the circle, we can obtain a function for any other circle passing through those two points.

What about circles that do not intersect at all? It turns out we can still find a function whose zero locus is the first circle, another function whose zero locus is the second circle, and make a linear combination of those two functions in order to obtain a function whose zero locus is another circle or even a straight line. But the zero locus of a linear combination of these functions cannot intersect the zero loci of either of the other two functions (except, of course in the case where one of the coefficients in the linear combination is zero).

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If you take the product of two polynomials rather than a linear combination, you get a zero locus that is simply the union of the zero loci of the original two polynomials. For example, take $f_4(x,y) = x + 2y$ and $f_5(x,y) = 2x - y$; the zero locus of each of these functions is a line, and the locus of $(x + 2y)(2x - y) = 0$ is a pair of intersecting lines, the union of the first two lines. If we now take yet another function whose zero locus is a line, $f_6(x,y) = x - 2,$ and subtract it, so we now have the polynomial $(x + 2y)(2x - y) - (x - 2),$ the zero locus of this new polynomial (where the polynomial is the function for a pair of straight lines, minus the function for one straight line) happens to be a hyperbola.

It does happen that the polynomial $y(y-15) - 4a(x-12)$ has a parabola as its zero locus, but not just because you can decompose the polynomial into the polynomials $y,$ $y-15,$ and $4a(x-12),$ so that the zero locus of $y(y-15)$ is a pair of lines and the zero locus of $4a(x-12)$ is a third line. It matters that the first pair of lines are parallel. We can see from the polynomial $y(y-15) - 4a(x-12)$ that whatever its zero locus is, that curve must pass through the intersection of $y=0$ and $4a(x-12) = 0$ and also pass through the intersection of $y-15=0$ and $4a(x-12) = 0,$ and not intersect any of those three lines at any other point; after examining all possible zero loci of a quadratic polynomial of two variables (the conic sections and their degenerate cases), the only one that fits these criteria is a parabola. We also know that the axis of the parabola is parallel to the two parallel lines. But we do not know where the vertex is merely by looking at the three lines. The vertex may be in many places depending on the value of $a,$ even though the three lines are the same in all of these cases.

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Now, just out of curiosity, let's consider these four equations of lines: \begin{align} 2x &= 0, \\ x - 2 &= 0, \\ x + y &= 0, \\ x - y &= 0. \end{align} Take the product of the first two polynomials, then subtract the product of the other two polynomials to obtain the polynomial $$ 2x(x - 2) - (x + y)(x - y). $$ The zero locus of this polynomial is a circle with center $(2,0)$ and radius $2.$

But if you take other linear combinations of $2x(x - 2)$ and $(x + y)(x - y)$ and examine the zero locus of the resulting polynomial, you may find that the locus is an ellipse, a parabola, or a hyperbola.

Moreover, when you obtain a parabola, the axis of the parabola is not parallel to the two parallel lines $2x = 0$ and $x - 2 = 0.$ It happens to be perpendicular in this case, but if (for example) we replace $x - y = 0$ by $x - 2y = 0,$ we can get a parabola whose crosses the parallel lines at a different angle.

In short, you can tell a few things about a polynomial curve by decomposing the polynomial into polynomials of lines, but there are a lot of things you cannot determine just by looking at that set of lines.