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**KiWoonG****Guest**

Hi,,

i needed help for parametrics,,

here's a sample question,,,

A parabola has parametric equations x=8t and y=4t^2

Find the coordinates of its focus and equation of its directix.

can somebody help me on this??

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I'm not sure what you mean by focus and directix, but it's easy enough to convert it into a normal cartesian equation.

x=8t

t=x/8

y=4t²

y=4(x/8)²

y=4(x²/64)

y=x²/16

And from that, you should be able to get any information you want.

Why did the vector cross the road?

It wanted to be normal.

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**ajp3****Member**- Registered: 2005-08-25
- Posts: 9

use the focus directrix form of the eqn of a parabola:

x² = 4py (or y² = 4px for the horizontal case); note that p = the distance from the vertex to the focus = the distance from the vertex to the directrix

now, mathsyperson has already done most of the work!

take x² = 16y = 4(4y), so we see p = 4.

this means our parabola (with vertex (0,0) ) has focus (0, 4) and directrix y = -4

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