|
|
|
|
|Store|
|
|
|
|
CTK Exchange
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.
|
|
|
Alert | IP |
Printer-friendly page |
Reply |
Reply With Quote | Top |
|
|
You may be curious to have a look at the old CTK Exchange archive.
Please do not post there.
|Front page|
|Contents|
Copyright © 1996-2018 Alexander Bogomolny
|
|
Printer-friendly copy
Email this topic to a friend