Can't teach here. Try:

http://www.mathwords.com/f/focus_parabola.htm

http://www.purplemath.com/modules/parabola.htm

- Tue Mar 25, 2014 10:57 am
- Forum: Advanced Algebra ("pre-calculus")
- Topic: Parabola Problem
- Replies:
**2** - Views:
**1252**

- Tue Mar 25, 2014 10:52 am
- Forum: Intermediate Algebra
- Topic: Arch Problem
- Replies:
**2** - Views:
**865**

No graph paper?

- Mon Mar 17, 2014 5:27 am
- Forum: Advanced Algebra ("pre-calculus")
- Topic: Logarithm equation
- Replies:
**5** - Views:
**1476**

JC531 wrote:log_{5}(x-1)

Did you know that equals log(x-1) / log(5) ?

- Sun Mar 16, 2014 7:31 pm
- Forum: Geometry
- Topic: Parabola equation
- Replies:
**2** - Views:
**1900**

I didn't times it all out, but I got y = [(e-v)/(d-u)^2](x-u)^2 + v I think mine probably ends up like yours if you do all the steps. (Do you have to do all that?) Ok, but a parabola equation is of this format: y = Ax2 + Bx + C I really should have made mine clearer: y = Ax^2 + Bx + C where: A = (e...

- Sun Mar 16, 2014 4:53 am
- Forum: Geometry
- Topic: Parabola equation
- Replies:
**2** - Views:
**1900**

GIVEN:

coordinates of vertex: (u,v)

coordinates of a point on parabola: (d,e)

Find equation of parabola

I got:

y = [(e - v) / (d^2 + u^2 - 2du)]x^2 - (2au)x + au^2 + v

Can someone confirm? Thank you.

coordinates of vertex: (u,v)

coordinates of a point on parabola: (d,e)

Find equation of parabola

I got:

y = [(e - v) / (d^2 + u^2 - 2du)]x^2 - (2au)x + au^2 + v

Can someone confirm? Thank you.

Any reasons why this forum is so inactive?

I joined yesterday, and there's been no posts since.

And most "last posts" in categories are earlier than 2014.

Kind of disappointing, given all the wonderful lessons available here...

I joined yesterday, and there's been no posts since.

And most "last posts" in categories are earlier than 2014.

Kind of disappointing, given all the wonderful lessons available here...