Check out some similar questions!

akotoh · Jun 28, 2009, 06:16 AM

hello... :) I hope u can help me with this.. I find it very difficult for me bcoz our prof gives us example but this problems are not d sme in any of his example. :( I've tried my best to answer this problem but still n0thing happens.. I hope u can really help me..

1.) x^2-2x-y=0
I thot diz will become:
x^2=2x+y

2.) y^2+3x-2y+7=0
y^2=-3x+2y-7

3.) 2y^2-5x+3y-7=0
2y^2=5x-3y+7

galactus · Jun 28, 2009, 07:38 AM

With this? Becoz? What language is that?

Unknown008 · Jun 28, 2009, 08:53 AM

Ah, the youngsters of these days...

I think that you need to put in the form of

$(y-k)^2 = 4p(x-h)$

Am I right? I just have found some questions about those, but haven't done it at school yet.

akotoh · Jun 28, 2009, 04:57 PM

Mr. Galactum.. please help me in this problem.. :(

akotoh · Jun 28, 2009, 05:09 PM

Sorry for the spelling mr. galactum.. that should be: "with this" and "because".

Unknown008 · Jun 29, 2009, 08:24 AM

*Sigh* it's galactus, akotoh, not galactum

galactus · Jun 29, 2009, 03:45 PM

The vertex is halfway between the focus and the directrix. The distance from the focus to

the vertex or from the vertex to the directrix is normally indicated by 'p'.

If the parabola has an axis of symmetry parallel to the x-axis, then it has equation

$(y-k)^{2}=\pm 4p(x-h)$ . Where (h,k) are the coordinates of the vertex.

If it opens in the positive x direction, then it has a + sign. If it opens in the negative x direction, then it has a - sign.

If the parabola has an axis of symmetry parallel to the y-axis, then it has equation

$(x-h)^{2}=\pm 4p(y-k)$ .

If it opens up then it has a + sign and if it opens down, then it has a negative sign.

See if you can get them in the above forms.

1.) $x^{2}-2x-y=0$

You could rewrite this one as $y=x^{2}-2x$

It has focus at (1, -3/4)
Vertex at (1,-1)
Directrix at y=-5/4

The vertex can be found by $\frac{-b}{2a}$ In this case, a=1 and b=-2

$x=\frac{-(-2)}{2(1)}=1$ . Sub back into the equation and get y=-1

2.) $y^{2}+3x-2y+7=0$

You could write this one as $x=\frac{-1}{3}y^{2}+\frac{2}{3}y-\frac{7}{3}$

This one opens to the left. It has vertex at (-2,1), Focus at (11/4,1) and directrix at x=-5/4

3.) $2y^{2}-5x+3y-7=0$

Opens to the right.

$x=\frac{2}{5}y^{2}+\frac{3}{5}y-\frac{7}{5}$

Vertex at (-13/8, -3/4), Focus at (-1,-3/4), Directrix at x = -9/4