Question

Deriving the equation of a parabola,
(x-4)^2+y^2-16y+16=y^2+4y^2+4
Been stuck on this forever.

Answer 1

what on earth is that?

Answer 2

I'm supposed to simplify it

Answer 3

\[(x-4)^2+y^2-16y+16=y^2+4y^2+4 \] combine like terms on the right and get
\[(x-4)^2+y^2-16y+16=5y^2+4 \]

Answer 4

then subtract if you want to set it equal to zero and get \[(x-4)^2-4y^2-16y+12=0\] if you like
it is not a parabola for sure

Answer 5

really

Answer 6

it is a hyperbola
was that the original question?

Answer 7

one sec

Answer 8

ok so

Answer 9

it started off as two distance formulas

Answer 10

im try to find the distance between the directx and the focus

Answer 11

the directx being 0,2

Answer 12

the focus being 4,4

Answer 13

the directrix of a parabola is not a point, it is a line

Answer 14

I know the answer but i need to show my work and the problem says to set up the problem as two distance formulas being equal to each other and to simplify the equation

Answer 15

is it the line \(y=2\)? it cannot be a point \((0,2)\) although that could be a point on the line

Answer 16

yes

Answer 17

the distance between the line \(y=2\) and the point \((4,4)\) does not use the distance formula
that distance is \(2\)

Answer 18

i mean -2

Answer 19

not 2

Answer 20

how far down from the point \((4,4)\) is the line \(y=-2\)?

Answer 21

6 units right?

Answer 22

yes

Answer 23

http://hillsborough.flvs.net/educator/student/frame_toolbar.cgi?tblackburn6*eternalraze*mpos=1&spos=0&option=hidemenu&slt=o7bg2KUMS0PMg*3940*http://hillsborough.flvs.net/webdav/educator_geometry_v15/index.htm
this is the lesson

Answer 24

lesson 9.05 page 3

Answer 25

The link doesn't work, but you can represent the directrix as a point if you write it as
(x , 2)

Answer 26

Ok so im supposed to explain how to find the equation of the parabola given the focal point and the directrix.

Answer 27

\[4p(y-k)^2=(x-h)^2\] where the vertex is \((h,k)\) and the distance between the focus and then vertex (which is equal to the distance between the vertex and the directrix, which is half the distance between the focus and the directix) is \(p\)

Answer 28

that was wrong
sorry it is
\[4p(y-k)=(x-h)^2\]

Answer 29

so if the focus is \((4,4)\) and the directrix is \(y=-2\) then the vertex is \((4,1)\) half way between them

Answer 30

the distance between \((4,1)\) and \((4,4)\) is \(3\) so the equation is
\[4\times 3(y-1)=(x-4)^2\]

Answer 31

here is the check
note the focus and the directrix
http://www.wolframalpha.com/input/?i=parabola+12%28y-1%29%3D%28x-4%29^2

Answer 32

@satellite73, @EternalRaze said earlier that the problem says to set up the problem as two distance formulas being equal to each other and to simplify the equation.

Answer 33

ick
no one used the distance formula to find a vertical distance
that is using pythagoras, but vertical distance doesn't use it
the distance between \((a, b)\) and \((a,c)\) is \(|b-c|\)

Answer 34

@satellite73, if someone told me I had to solve x^2 + 5x + 6 using the quadratic formula, I would have to use the quadratic formula rather than factoring even though I prefer factoring better.