Find the standard form of the equation of the parabola with a focus at (0, 8) and a directrix at y = -8.

Question

anonymous · Answer

y = 1/32 ^x2

y^2 = 8x

y^2 = 32x

y = 1/8 ^x2

anonymous · Answer

@amistre64 Could you help me with this?

amistre64 · Answer

your notations seem a little off in the options

amistre64 · Answer

but the mechanics involved is such that the definition of a parabola is all the points (x,y) that are the same distance from the focus (a,b) and some line directix ... in this case y=k

the distance formula is therefore a good formula to employ if need be:

df^2 = (x-a)^2 + (y-b)^2

and

dd^2 = (x-x)^2 + (y-k)^2 

equating them we get:

df^2 = dd^2

(x-a)^2 + (y-b)^2 = (x-x)^2 + (y-k)^2 

and simplify with algebra

(x-a)^2 + (y-b)^2 = (y-k)^2 

(x-a)^2 + y^2 -2by +b^2 = y^2 -2k y +k^2 

(x-a)^2 -2by +b^2 = -2k y +k^2

(x-a)^2 +b^2 - k^2 = 2by -2k y

(x-a)^2 +b^2 - k^2 = (2b -2k) y


(x-a)^2 +b^2 - k^2
------------------  =  y
        2 (b-k)

amistre64 · Answer

lets see if the values make this prettier lol

amistre64 · Answer

(a,b) = (0,8);  k=-8


(x-0)^2 +8^2 - 8^2
------------------  =  y
        2 (8+8)


  x^2
------  =  y
 2 (16)

anonymous · Answer

Sorry really slow internet, sigh. So it would be the third answer choice? (y^2= 32x)

anonymous · Answer

Wait no it's y = 1/32 x2?

amistre64 · Answer

the site is definitely not working well at the moment,  but yes:  y = 1/32  x^2  fits

anonymous · Answer

I thought it only happened to me, lol. THANKS SO MUCH!