Question

what is the equation of the quadratic graph with a focus of (3,6) and a directrix of y=4???

Answer 1

@phi I don't understand what I am supposed to do. am I supposed to plug (3,6) into the vertex form y=a(x-h)^2+k

Answer 2

I guess they want you to learn that the vertex is the point exactly in between the focus and the directrix.

I always plot the focus and the directrix to make it more clear what is going on.

Answer 3

can you find the vertex ?

Answer 4

the other thing you need to know is "a" in
y=a(x-h)^2+k

is equal to \( \frac{1}{4p} \)
"p" is the distance from the vertex to the focus (or the distance from the vertex to the directrix or ½ the distance between the focus and the directrix)

Answer 5

Here is a plot

Answer 6

so should i plug the numbers into the distance formula

Answer 7

you don't need to. the shortest distance from a point to a line is a line that makes a right angle with the given line. in other words, the distance is the distance from the point *straight down* to the directrix

Answer 8

so its just this??

Answer 9

point B is not the closest point on the line y=4
y=4 is a line that goes horizontal.
go *down* (not sideways or diagonally)

Answer 10

yes. If you also plot the directrix y=4
it would look nice.

the VERTEX will be on that line, exactly ½ between the 2 points.

Answer 11

but the question says " what us the equation of the quadratic graph with a focus of (3,6) and directrix of y=4? I have the graph but i don't understand how to put it into an equation.

Answer 12

The first step to answering the question is to find the vertex.

The vertex is exactly half way between points A and B in your plot.

can you find the x,y coordinates of the vertex?

Answer 13

(3,5)

Answer 14

yes, the vertex is at (3,5)
that means you know h and k for your formula

y = a(x-h)^2 + k

Answer 15

remember (h,k) is the vertex. h is 3 and k is 5 for this problem

Answer 16

what is a

Answer 17

the other thing you need to know is "a" in
y=a(x-h)^2+k

"a" is equal to \( \frac{1}{4p} \)
"p" is the distance from the vertex to the focus (or the distance from the vertex to the directrix or ½ the distance between the focus and the directrix)

Answer 18

so a is 2

Answer 19

or 1

Answer 20

Someone help me!!

Answer 21

Okay

Answer 22

First what is "p" ?

Answer 23

Which of the following does not describe the dependent variable?
A.
It is the range.
B.
It is the y-values.
C.
It is the output.
D.
It is the x-values.

\

Answer 24

"p" is the distance from the vertex to the focus
the vertex is (3,5) and the focus is (3,6)

Answer 25

so you ain't gonna help me?

Answer 26

in a minute. first go make your own question. I am in the middle of getting help my self.

Answer 27

you do not need to post a picture.
what is the distance between point C (the vertex) and point A (the focus) ?

Answer 28

Which values represent the independent variable?

(–2, 4), (3, –2), (1, 0), (5, 5)
A.
{–2, 3, 1, 5}
B.
{4, –2, 0, 5}
C.
{–2, 4, 3, –2}
D.
{–2, –1, 0, 5}

Answer 29

1

Answer 30

you could use the distance formula
\[ \sqrt{(3-3)^2 + (6-5)^2 } = \sqrt{0+1}= 1 \]

Answer 31

now use
a= 1/(4*p)

Answer 32

Here is what your parabola looks like

Answer 33

so its .25 or 1/4

Answer 34

yes. now you have the full equation

Answer 35

if you type that equation into geogebra, it will plot it.

Answer 36

y=1/4(x-3)^2+5

Answer 37

yes

Answer 38

do you know how to change the geogebra graph to have all the lines on the coordinate plane

Answer 39

The right most button up top. if you select it, it will move the center of the graph.
If you click on its lower right corner (where the tiny triangle is) you can select zoom out.