An archway will be constructed over a walkway. A piece of wood will need to be curved to match a parabola. Explain to Maurice how to find the equation of the parabola given the focal point and the directrix (8).

d = (square root) (Xv2 - 3)^2 + (Yv2 - 6)^2
d = (square root) (Yv2 - 8)^2

I need to something like -1/-8x^2 + 5/4x + 23/8 = y
(example given by school)

Its to do an arc.

I got to Xv2^2 -6Xv2 + 9 + Yv2^2 -8Yv2 + 16 = ???

Question

An archway will be constructed over a walkway. A piece of wood will need to be curved to match a parabola. Explain to Maurice how to find the equation of the parabola given the focal point and the directrix (8).

d = (square root) (Xv2 - 3)^2 + (Yv2 - 6)^2
d = (square root) (Yv2 - 8)^2

I need to something like -1/-8x^2 + 5/4x + 23/8 = y
(example given by school)

Its to do an arc.

I got to Xv2^2 -6Xv2 + 9 + Yv2^2 -8Yv2 + 16 = ???

anonymous · Answer

You are forgetting the weird exponent that is below the number. I don't know if that effects the equation at all though.

What I'm supposed to get is something like $$-\frac{ 1 }{ 8 } x^2 + \frac{ 5 }{ 4 } x + \frac{ 23 }{ 8 } = y$$

Which someone how becomes this
-0.13x^2 + 1.25x + 2.88

(Not my problem, its their example)

anonymous · Answer

Expand it?

And give me a second, I'll stitch together some screenshots of what they gave for instructions.

anonymous · Answer

attachment

anonymous · Answer

I don't really understand any of this as I was horrible at algebra. I need it like what I said before, "-0.13x^2 + 1.25x + 2.88" so I can put it in stupid geogebra to submit for my project.

anonymous · Answer

Because I have to put it into geogebra, or I don't get a grade.

campbell_st · Answer

so let's check... the focus is at (3, 6) and directrix is at y = 8

anonymous · Answer

yes

campbell_st · Answer

great so using the distance formula method

pick a point P(x, y) on the parabola

distance Focus to P

$$d = \sqrt{(x - 3)^2 + (y - 6)^2}$$

P to the directrix

$$d = \sqrt{x -x)^2 + (y - 8)^2}$$

equating and squaring you get

$$(x -3)^2 + (y - 6)^2 = (y - 8)^2$$

now subtract (y - 6)^2 from both sides

$$(x -3)^2 = (y -8)^2 - (y -6)^2 $$

does that make sense..?

campbell_st · Answer

so simplify the right hand side

$$(x-3)^2 = y^2 - 16y + 64 - y^2 + 12y - 36$$

which becomes 

$$(x -3)^2 = -4y + 28$$

factor the right hand side

$$(x -3)^2 = -4(y - 7)$$

does that make sense...?

campbell_st · Answer

make y the subject 

$$-\frac{(x -3)^2}{4} + 7 = y$$

you can enter the equation as its written above into Geogebra...

campbell_st · Answer

it will then write the equation in expanded form with decimal coefficients...

campbell_st · Answer

and I just graphed the parabola in geogebra and it worked perfectly...

anonymous · Answer

how do you graph it in it?

campbell_st · Answer

just enter the equation I've written above 

$$y = -\frac{(x -3)^2}{4} + 7$$

anonymous · Answer

y = -0.25x^2 + 1.5x + 4.75?