For the parabola $(y-3)^2 = 6(x+2)$, what are the vertex-to-focus, vertex-to-directrix, and focus-to-directrix distances?

Question

I am trying to get ahead in a class and I don't know how to answer the following questions:

What is the distance from the vertex to the focus for the parabola $$(y-3)^2 = 6(x+2)$$ What is the shortest distance from the vertex to the directrix for this parabola?

What is the shortest distance from the focus to the directrix for this parabola?

Answer 1

Vertex lies at the point $$(-2,3)$$ and focus lies at the point $(-\dfrac{1}{2}, 3)$ and equation of directrix is, $x=-7/2$

Now the distance between vertex and focus will be equal to $$-\dfrac12-(-2)=\dfrac32$$

The distance between vertex and directrix will be equal to $$-2-(-7/2)=3/2$$

The distance between focus and directrix will be equal to $$3/2+3/2=3$$

Answer 2

Let's look at a more general case first.

The distance from the focus to each point on the parabola equals the distance from that point to the directrix. And this distance is minimized at the vertex.

Let's start with a parabola with focus $(p,0)$ and directrix $y = -p$

$p$ is the distance from focus to vertex, and from vertex to directrix.

The points on the parabola are the $(x,y)$ pairs such that:

$\sqrt {(x-p)^2 + y^2} = x+p$

Squaring both sides and expanding the binomials

$x^2 -2px + p^2 + y^2 = x^2 + 2px + p^2\\ y^2 = 4px$

That last line is the key to this whole problem.

Tranlating the parabola $(y-3)^2 = 4p(x+2)$ does nothing to change $p$

Now, it is quite likely that you have already derived all of this already. And, it would be fine to start here:

$4p = 6\\ p = \frac 32$