1. ## Parabolas

Ok, I've been given this:

$y^2=6x$

and I have to find the vertex, focus, and directrix of this graph.

I know how to find the vertex. In this case, it is at the origin. However, I don't have the slightest idea how to find the focus or the directrix.

Help would be appreciated here

2. Originally Posted by mathgeek777
Ok, I've been given this:

$y^2=6x$

and I have to find the vertex, focus, and directrix of this graph.

I know how to find the vertex. In this case, it is at the origin. However, I don't have the slightest idea how to find the focus or the directrix.

Help would be appreciated here
The first thing to do is write the equation as $x = \frac{y^2}{6}$. Then you have an equation in the form $x = a(y - y_0)^2 + x_0$. You correctly identified that the vertex is (0, 0). Now, the focus and directrix are both a distance of $\frac{1}{4a}$ from the vertex.

Hence:

Focus is $(\frac{3}{2}, 0)$

Directrix is $x = -\frac{3}{2}$

3. That's not what the back of the book says.

The back of the book says that the focus is $(-\frac{3}{2}, 0)$ and the directrix is $x=-\frac{3}{2}$

4. Originally Posted by mathgeek777
That's not what the back of the book says.

The back of the book says that the focus is $(-\frac{3}{2}, 0)$ and the directrix is $x=-\frac{3}{2}$
I fixed my formula

5. Originally Posted by icemanfan
Now, the focus and directrix are both a distance of $\frac{1}{4a}$
I still fail to see how that relates to the finding of the focus. I can figure out one of the terms of the focus just by looking at which direction the parabola is going, but as far as finding the other term, I am still lost.

6. Originally Posted by mathgeek777
I still fail to see how that relates to the finding of the focus. I can figure out one of the terms of the focus just by looking at which direction the parabola is going, but as far as finding the other term, I am still lost.
The focus should be "inside" the parabola. Since we have a parabola that opens to the right, the focus should be to the right of the vertex. And its distance from the vertex is $\frac{1}{4a}$, with $a = \frac{1}{6}$.

The directrix should never touch the parabola and a perpendicular drawn to the directrix from the vertex will have length $\frac{1}{4a}$, which is why we must have the line $x = -\frac{3}{2}$.

Edit: Your book has the wrong focus.