(Help?)
Find the equation of the parabola with focus (2, 3) and directrix y = -1.

A) (y - 1)^2 = 8(x - 2)
B) (y - 1)^2 = 4(x - 2)
C) (x - 2)^2 = 8(y - 1)
D) (x - 2)^2 = 4(y - 1)

Question

(Help?) 	
Find the equation of the parabola with focus (2, 3) and directrix y = -1.

A) (y - 1)^2 = 8(x - 2)	
B) (y - 1)^2 = 4(x - 2)	
C) (x - 2)^2 = 8(y - 1)	
D) (x - 2)^2 = 4(y - 1)

anonymous · Answer

$$
(x-2)^2=8 (y-1)
$$
I will tell you why below

anonymous · Answer

please do explain.

anonymous · Answer

The parabola is the set of points equidistant from the focus and directirx

anonymous · Answer

$$
(y+1)^2=(x-2)^2+(y-3)^2

$$

anonymous · Answer

Expand and put in the right form.

anonymous · Answer

$$
y^2+2 y+1=x^2-4 x+y^2-6 y+13

$$

anonymous · Answer

$$
x^2-4 x+y^2-\left(y^2+2 y+1\right)-6 y+13=x^2-4 x-8 y+12=0
$$

anonymous · Answer

did you distribute.

anonymous · Answer

$$
x^2 - 4 x - 8 y + 12 = 
 x^2 - 4 x + 4 + 8 - 8 y = (x - 2)^2 + 8 - 8 y = 0\
(x - 2)^2 =  8 y-8=8(y-1)
$$

anonymous · Answer

ah never mind i see what you did. That makes tons of sense now to me. Thanks a lot! :)

anonymous · Answer

YW