Question

Question 1
The vertex form of the equation of a vertical parabola is given by , where (h, k) is the vertex of the parabola and the absolute value of p is the distance from the vertex to the focus, which is also the distance from the vertex to the directrix. You will use the GeoGebra geometry tool to create a vertical parabola and write the vertex form of its equation. Open GeoGebra, and complete each step below. If you need help, follow these instructions for using GeoGebra.

Part A
Mark the focus of the parabola you are going to create at F(6, 4). Draw a horizontal line that is 6 units below the focus. This line will be the directrix of your parabola. What is the equation of the line?

Answer 1

I NEED HELP!!!

Answer 2

I don't understand what it means by what is the equation of the line

Answer 3

In order to find the focus of a parabola, you must know the equation of a parabola.
The vertex form is y=a(x−h)2+k

Answer 4

That makes no sense to meeee

Answer 5

I need like in depth explanation of how this shi works tbh

Answer 6

Sorry for cussing I have been sitting on this problem for like 2 hours and it's just frustrating me

Answer 7

hold on let me try to figure it out

Answer 8

okie ty so much

Answer 9

Let S(x,y) be any point on the parabola.
ST

Answer 10

\[\because ~directrix~is~ 6~units~below~focus ~F(6,4)\]
so 4-6=-2
eq. of directrix is y=-2
or
y+2=0
Let ST perpendicular directrix .
\[ST=\frac{ |y+2| }{ \sqrt{0^2+1^2} }=|y+2|\]
distance between (x,y) and (6,4) is
\[SC=\sqrt{(x-6)^2+(y-4)^2}\]
ST=SC
\[ST^2=SC^2\]
\[(y+2)^2=(x-6)^2+(y-4)^2\]
\[y^2+4y+4=(x-6)^2+y^2-8y+16\]
\[y^2+4y+4-y^2+8y-16=(x-6)^2\]
\[12y-12=(x-6)^2\]
or
\[(x-6)^2=12(y-1)\]