Equation of a Parabola: Can Any Point on the Graph Satisfy the General Equation?

  • Thread starter Thread starter kevinshen18
  • Start date Start date
  • Tags Tags
    Parabola
kevinshen18
Messages
23
Reaction score
0
The general equation of a parabola is:
(y - y1) = A(x - x1)^2

A is the scaling factor
y1 is the y coordinate of a vertex/point on the parabola
x1 is the x coordinate of a vertex/point on the parabola

My two questions are:
1. Can (x,y) be any point on the graph ?
2. If so, then if x - x1 = 0 and y- y1 = 0 (if the difference between the points are 0) then does that mean the point (x,y) is on the parabola?
 
Mathematics news on Phys.org
kevinshen18 said:
The general equation of a parabola is:
(y - y1) = A(x - x1)^2

A is the scaling factor
y1 is the y coordinate of a vertex/point on the parabola
x1 is the x coordinate of a vertex/point on the parabola

My two questions are:
1. Can (x,y) be any point on the graph ?
Yes, (x, y) refers to any point on the parabola.

2. If so, then if x - x1 = 0 and y- y1 = 0 (if the difference between the points are 0) then does that mean the point (x,y) is on the parabola?
Your question is a little confusing (or confused). Again, (x, y) can refer to any point on the parabola. "if x- x1= 0 and y- y1= 0" then x= x1 and y= y1 is a specific point on the parabola. In fact, it is the vertex of the parabola- the lowest point if A is positive, highest point if A is negative.
 
kevinshen18 said:
The general equation of a parabola is:
(y - y1) = A(x - x1)^2

A is the scaling factor
y1 is the y coordinate of a vertex/point on the parabola
x1 is the x coordinate of a vertex/point on the parabola

My two questions are:
1. Can (x,y) be any point on the graph ?

That's what the equations of these curves are for.

Instead of having to compile a list of the infinite number of point coordinates which fall on the curve, a simple equation can be used to generate one point or many points, all of which will fall on the curve

2. If so, then if x - x1 = 0 and y- y1 = 0 (if the difference between the points are 0) then does that mean the point (x,y) is on the parabola?

Any point (x,y) which satisfies the equation of the parabola is a point on that parabola.

Now, why all of this confusion?
 
I would say that (x,y) is some point in the x,y plane, may on the parabola maybe not.

Points on a parabola are given by: (x, (x-x1)2+ y1))
 
In the original post the reference was to (x, y) satisfying the equation (y- y_1)= A(x- x_1)^2.

Those (x, y) are points on the parabola, not arbitrary points in the plane.
 
Thanks guys. So if (x,y) = (x1,y1), then this satisfies the parabola equation and (x1,y1) is the vertex?
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top