Distance from vertex to focus in a parabola

[John 123]

Homework Statement

I am a bit rusty on parabolae.
I am doing a question on projectiles and have found the coordinates of the vertex of a parabola as:
$$(\frac{(v_0)^2\sin\alpha\cos\alpha}{g},\frac{(v_0)^2(\sin^2\alpha)}{2g})$$


The question now requires you to show that the distance from the vertex to the focus is given by:
$$\frac{(v_0)^2(\cos^2\alpha)}{2g}$$

and that the equation of the directrix is
$$y=\frac{(v_0)^2}{2g}$$

John

Homework Equations

The Attempt at a Solution

 

[Mentallic]

The distance from the vertex to the focus is equal to the distance from the vertex to the directrix. So all you have to do is find the perpendicular distance between the directrix and vertex (thus, the directrix is going to be vertically above the vertex and focus in this case).

 

[John 123]

Many thanks Mentallic
My apologies but I did not include the original equation in my original post.
However, I have now solved the problem by converting the original equation by completing the square etc into the form:
$$(x-h)^2=4p(y-k)$$
where (h,k) are the vertex coordinates and p is the distance from the vertex to the focus.
I was then able to find p and then the equation of the directrix.
John

 

[Mentallic]

Ahh hehe I was going to say! For a second there I thought you were going to take a very complicated and winding road when all that needs to be done is simple distance calculations