# Math Help - Parabolas

1. ## Parabolas

I was given this parabola y = 2x^2 + 12x + 17

Can anyone change this to the "a(x-h)^2 + k" standard form?

I also need to find the vertex,focus,directrix and length of the latus rectum of this parabola.

How would I go about doing that? Any help on this would be much appreciated..

2. Originally Posted by 14041471
Can anyone change this to the "a(x-h)^2 + k" standard form?
$y = 2(x^2 + 6x) + 17
$

$y = 2(x^2 + 6x + 9) + 17 - 18$

$y = 2(x + 3)^2 - 1$

3. ## thankyou

thanks alot, please could you help me find the latus ractum.

4. Originally Posted by 14041471
thanks alot, please could you help me find the latus ractum.
Any replies at http://www.mathhelpforum.com/math-he...parabolas.html can be used to answer some of the remaining questions here.

5. ## the focus of a parabola

I believe the focus is (3,-7/8) am i correct? if no what is the awnser

I also have to find the diretrix and have the horizontal line (-9/8) is that correct?

6. What I wrote before, if you saw it, was wrong! I was thinking "directrix", not "latus rectum". Strictly speaking the latus rectum is the line segment from one point on the parabola to another point on the parabola, perpendicular to the axis and through the focus. Some texts, however, also use "latus rectum" to refer to the length of that line segment. You had better check which your textbook or teacher intends.

With your example, $y= 2(x+ 3)^2- 1$ the axis is vertical so the latus rectum will be horizontal: between two points on the parabola having the same y value as the focus. After you have found the focus, say $\left(x_f, y_f\right)$, you will need to solve $2(x+3)^2- 1= y_f$. That will have two solutions. The latus rectum, as a line segment will be the line segment haveing those x coordinates and [tex]y_f[/itex] as y coordinate. As a single number, the lenght of that segment, it is the difference between those two x values.