Algebraically, construct a line that goes through the vertex, the focus, and the discriminant

Question

parthkohli · Answer

I'm not sure if we have a point called the "discriminant".

ahsome · Answer

Like this:|dw:1411386997220:dw|

ahsome · Answer

@ParthKohli , help

ahsome · Answer

I would be guessing we would need an equation equal to:
$$y=mx+c$$

parthkohli · Answer

If you want to construct a straight line algebraically, it's simple: you've got to know the location of those two points.

ahsome · Answer

It is a straight line

ahsome · Answer

I just suck at drawing ;)

anonymous · Answer

lol what you want , I didn't get

ahsome · Answer

I need to find, algebraicly, how to construct a line that cuts through the vertex, the focus AND the dirictrix of a parabola. The line is straight.

parthkohli · Answer

I believe the equation is$$x = -b/2a$$

ahsome · Answer

I think we would have to use the equation:
$$y=mx+c$$
Since we know there is no gradient, $m=1$, right?
And since it is straight (And I know that it cuts the y axis at 4), the $c$ value is 4?

ahsome · Answer

I know the value of vertex. (4,4)

parthkohli · Answer

Have you been specifically given a parabola?

ahsome · Answer

Yes. its formula is:
$$x=0.6(y-4)^2+4$$
Or
$$y=\pm\sqrt{\frac{x-4}{0.6}}+4$$

ahsome · Answer

@ParthKohli?

ahsome · Answer

@No.name @iGreen Help!

ahsome · Answer

Help @perl !

ahsome · Answer

@iGreen HELP!

igreen · Answer

I have no idea..sorry :(

ahsome · Answer

Oh :(
Do you know anybody that will @iGreen ?

igreen · Answer

No..sorry. I know one person, but he's offline.. (@phi )

perl · Answer

what would you like help with

perl · Answer

i think theres a typo in your question, it should say, go through focus, vertex and directrix

perl · Answer

not discriminant