How Do You Find the Vertex and Axis of Symmetry in a Parabola Equation?

  • Thread starter Thread starter Alain12345
  • Start date Start date
  • Tags Tags
    Conics Parabola
AI Thread Summary
To find the vertex of the parabola given by the equation (y-2)² = 4(x-3), identify the vertex at the point (3, 2). The axis of symmetry is the line x = 3, indicating that the parabola opens horizontally. The positive right-hand side of the equation confirms that the parabola opens to the right. The distance p, which is 1, represents the distance from the vertex to the focus. Understanding these elements allows for a clear analysis of the parabola's properties.
Alain12345
Messages
37
Reaction score
0
If I am given (y-2)2= 4 (x-3), how would I find the vertex, equation of the axis of symmetry, and the direction of the opening? I'm guessing that I have to use PF=PD, but it's confusing because it looks different from other things that I have done involving PF=PD.

Thanks.
 
Physics news on Phys.org
the right hand side of your equation is positive right?, what is the minimum value of x for which this is posible? and the corresponding value of y? this is the vertex. Now rebember that p=1 is the distance from the vertex to the focus and that you have and horizontal parabola.
 

Thread 'Finding the number of ways to arrange identical balls in a circle (3 different colors)'

I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
View full post »

Thread 'Making the denominator of a certain fraction real'

Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
View full post »

Similar threads

Downward parabola: value of p = 1/a?
Replies
1
Views
2K
B What is the focus and parameter of a parabola with vertex off the origin?
Replies
44
Views
4K
Arriving at parabola formula via distance formula
Replies
4
Views
2K
Finding the Axis of Symmetry in a Parabola: How to Solve the Problem
Replies
9
Views
2K
I can really use a hand on finding vertex
Replies
8
Views
2K
Identifying the equation of a parabola
Replies
2
Views
2K
Find the range of values of a + b
Replies
6
Views
1K
Parabola: Vertex =(?,0) and know 2 arbitrary points. How solve x?
Replies
12
Views
2K
Finding the focus of a parabola given an equation
Replies
4
Views
2K
Conic Parabolas in General Form
Replies
1
Views
2K

Hot Threads

Recent Insights

Back
Top