Question

Derive the equation of the parabola with a focus at (2, 4) and a directrix of y = 8.

a. f(x) = -1/8 (x - 2)2 + 6
b. f(x) = 1/8 (x - 2)2 + 6
c. f(x) = -1/8 (x + 2)2 + 8
d. f(x) = 1/8 (x + 2)2 + 8

Answer 1

Hello,


`y=ax^2+b+c\\ \Delta=b^2-4ac\\ Focus=(- (b)/(2a), (1-\Delta)/(4a) ) \\ Directrix=y=- (1+\Delta)/(4a) \\ - (b)/(2a)=2==\textgreater\ b=-4a\\ (1-\Delta)/(4a)=4==\textgreater\ \Delta=1-16a\\ - (1+\Delta)/(4a)=8==\textgreater\ \Delta=-1-32a\\ 1-16 a=-1-32a==\textgreater\ a=- (1)/(8) \ ,\ b=-4a= (1)/(2) \\ \Delta=-32a-1=3\\ b^2-4ac=3= (1)/(4)+ (1\ c)/(2) ==\textgreater\ c= (11)/(2) \\ y=- (1)/(8)x^2+ (1)/(2) x+ (11)/(2) \\`


`\boxed{y=- (1)/(8)(x-2)^2+6}`

Vertex is in the middle of directrix and focus on the axis of symmetry.

Answer A

Answer 2

In this problem, given the focus at (2,4) and directrix at y = 8. then it is implied that the parabola is facing downwards. The vertex hence is at the middle of the focus and the directrix, hence at (2, 6). The general formula of the parabola is y-k = -4a ( x-h)^2. Substituting, y -6 = -1/8 *(x-2)^2. Answer is A.