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
$\sqrt{(x-6)^2+(y-2)^2} = \frac{y-1}{ \sqrt{1^2} }$
 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 $(6,2-\dfrac{1}{2})=(6,\dfrac{3}{2})$. Hence, the parabola is: $(x-6)^2=2\cdot1\cdot(y-\dfrac{3}{2})\\\\x^2-12x+36=2y-3\\\\2y=x^2-12x+39\\\\\boxed{y=\frac{1}{2}x^2-6x+\dfrac{39}{2}}$ Now, derivating: $y'=\frac{1}{2}\cdot2x-6\\\\\boxed{y'=x-6}$