Question

Derive the equation of the parabola with a focus at (6, 2) and a directrix of y = 1.

Answer 1

let P(x,y) be any point on parabola
focus is S(6,2)
let M be the foot of perpendicular from P on directrix y=1 or y-1=0
then SP=PM

`√((x-6)^2+(y-2)^2) = (y-1)/( √(1^2) )`
squaring both sides
(x-6)^2+(y-2)^2=(y-1)^2
x^2-12x+36+y^2-4y+4=y^2-2y+1
x^2-12x+40=y^2-2y+1-y^2+4y
x^2-12x+40=2y+1
or 2y=x^2-12x+39
which eq of parabola

Answer 2

As the directrix is parallel to x-axis, the concavity of the parabola will be in vertical. Then, the general equation is in the form:
`(x-x_0)^2=2p(y-y_0)`, where
`(x_0,y_0)` are the coordinates of the vertice and p is the parameter (distance between the focus and the directrix). The parameter is
`p=1`. The vertice is the midpoint between the focus and the projection of the focus in the directrix. Then, the vertice is
`(6,2-(1)/(2))=(6,(3)/(2))`. Hence, the parabola is:
`(x-6)^2=2\cdot1\cdot(y-(3)/(2))\\\\x^2-12x+36=2y-3\\\\2y=x^2-12x+39\\\\\boxed{y=(1)/(2)x^2-6x+(39)/(2)}` Now, derivating:
`y'=(1)/(2)\cdot2x-6\\\\\boxed{y'=x-6}`