CTK Exchange Front Page Movie shortcuts Personal info Awards Reciprocal links Terms of use Privacy Policy Cut The Knot! MSET99 Talk Games & Puzzles Arithmetic/Algebra Geometry Probability Eye Opener Analog Gadgets Inventor's Paradox Did you know?... Proofs Math as Language Things Impossible My Logo Math Poll Other Math sit's Guest book News sit's Recommend this site    |Store|    CTK Exchange

 Subject: "Focus of parabola." Previous Topic | Next Topic
 Conferences The CTK Exchange College math Topic #527 Printer-friendly copy Email this topic to a friend Reading Topic #527
Ganitopasak guest
Aug-27-05, 07:40 PM (EST)

"Focus of parabola."

 Hello,How to calculate the focus of parabola given by:(x – 3)^2 = (y + 1)Thanks for any help.

delta59 guest
Aug-29-05, 07:27 AM (EST)

1. "RE: Focus of parabola."
In response to message #0

 The vertex of any parabola lies on the axis of symmetry and is equidistant from the focus and the directrice. In your case the vertex is V(3,-1). Assume focus F has coordinates F(3,-1+t) so directrice has equation y = -1 – t. Find any point on the parabola, for example A(2,0).Then AF^^2 = (2-3)^^2 + (0-(-1+t))^^2 = t^^2 – 2t + 2. And the distance from A to the directrice is t+1. So (t+1)^^2 = t^^2 – 2t + 2 .This gives t = 1/4.

 Conferences | Forums | Topics | Previous Topic | Next Topic
 Select another forum or conference Lobby The CTK Exchange (Conference)   |--Early math (Public)   |--Middle school (Public)   |--High school (Public)   |--College math (Public)   |--This and that (Public)   |--Guest book (Public) Educational Press (Conference)   |--No Child Left Behind (Public)   |--Math Wars (Public)   |--Mathematics and general education (Public) You may be curious to have a look at the old CTK Exchange archive.  