Math, parabolas and tangents

Today's word problem...

Find the equations of the tangent and normal lines to the parabola with the vertex at 0,3 and a focus at 0,0, where x=-1.

Hmmm I said to myself. No problem if i had the equation for the parabola...

I know what a vertex....But what is the focus....Time to do research.

Apparently a line drawn from the focus to any point on the parabola will equal the length of a line drawn from the same point on the parabola to the "directrix".

Geez.

Do not fear though! I know a vertex is the point of the parabola and I know the focus lies within it, so I know it opens down.

So I go about and read a tonne and work through some examples....

All I got to say is, what would be do without Pythagoras?

In this situation I know a central point (0,0) and I know a point on the parabola (x,y). I know the distance between these two points is equal to the distance from x,y to the directrix.

If I think enough I even know that. I have one point given to me on the parabola. The vertex. 0,3. The distance from the focus to the vertex? 3 units. Since the distance from the vertex to the directrix must equal this we know that it is also 3 units long, so it sits at 6.

So given that and given that pythag states A^2+b^2=c^2 we know have enough information to determine the formula.

A= the distance from the Focus X to the x on the parabola (x-0)
B= the distance from the Focus Y to the y on the parabola (y-0)
C= We know that when the vertex = 3 the distance here must also be 3 as it must equal the directrix. We have 6 units from the focus to the directrix and it opens down. As the point moves along the parabola the distance will get closer to 6. So we can safely say C = 6-y

We get

Sqrt((x-0)^2+(y-0)^2)=(6-y)^2

That works out to be y = -((x^2-36)/12)

Based on THAT formula we can then calculate a derivative to get us a formula to give us the slope of a tangent line at any point.
I get that to be y'=-24x/(-12)^2 -x/6

Nope STILL not finished yet.

We need the equation for the tangent and normal lines.

Turns out That Slope = rise/run.

Rise = y1-y2
Run = x1-x2
slope = m
so
m=(y1-y2)/(x1-x2)
OR
y1-y2=m(x1-x2)
We know an x co-ordinate, but we don't have a y. We can put the x at -1 back into the parabola formula and pick up a y value. I got -35/12 for the y.

We also need to know the slope at -1 so we put the -1 back into y' and get (-1)(-x)/6 = 1/6

Now we can go
y1+35/12=1/6(x+1)
and once it's worked out we get,
y=(x/6)+37/12
That gives us tangent.

Now the Normal.

Normal lines have a negative recipricol slope to the line of interest, in this case our tangent, which was 1/6. Our normal slope must therefor be -6/1 or -6

We put that back into our "point-slope formula"
y1+35/12=-6(x+1)
y=-6x-37/12

And that's everything.

parabola: y = -((x^2-36)/12)
Tangent at -1: y= (x/6)+37/12
Normal to Tangent at -1: y= -6x-37/12

I am sure they are not simplifed the prettiest, but hell.

If you go to http://gcalc.net/jar/gcalc3.jnlp and select graph plugin, you can input(ie copy and paste everything after the =) the functions and see how the all play out.