Question

Using a directrix of y = -2 and a focus of (1, 6), what quadratic function is created? a. f(x) = (1/8)(x-1)^2 - 2 b. f(x) = -(1/8)(x+1)^2 - 2 c. f(x) = -(1/16)(x+1)^2 - 2 d. f(x) = (1/16)(x-1)^2 + 2

Answer 1

`f(x)=1/16(x-1)^2+2to(d)`

Explanation:

`"from any point "(x,y)" on the parabola the focus and"`
`"directrix are equidistant"`

`"using the "color(blue)"distance formula"`

`sqrt((x-1)^2+(y-6)^2)=|y+2|`

`color(blue)"squaring both sides"`

`(x-1)^2+(y-6)^2=(y+2)^2`

`(y-6)^2-(y+2)^2=-(x-1)^2`

`cancel(y^2)-12y+36cancel(-y^2)-4y-4=-(x-1)^2`

`-16y+32=-(x-1)^2`

`-16y=-(x-1)^2-32`

`rArry=f(x)=1/16(x-1)^2+2to(d)`