Question

Driving equation of a parabola giving its focus and Directrix

Answer 1

a) Notice that Q is on the line x=-3, then the x-coordinate of Q is -3.

Since Q and P(x,y) are in the same horizontal line, then they must have the same y-coordinate, therefore:

$Q=(-3,y).$

b) Recall that the length of a segment is the distance between its terminal points, then:

$\begin{gathered} FP=\sqrt{(x-3)^2+(y-0)^2}=\sqrt{(x-3)^2+y^2}, \\ QP=\sqrt{(x+3)^2+(y-y)^2}=\sqrt{(x+3)^2}, \\ since\text{ }x>-3,\text{ }QP=x+3. \end{gathered}$

c) Recall that a parabola is a curve where any point is at an equal distance from a fixed point (the focus) and a fixed straight line (the directrix), then:

$FP=QP.$

d) Substituting FP and QP we get:

$x+3=\sqrt{(x-3)^2+y^2}^.$

Therefore:

$\begin{gathered} (x+3)^2=(x-3)^2+y^2, \\ x^2+6x+9=x^2-6x+9+y^2, \\ 12x=y^2, \\ x=\frac{1}{12}y^2. \end{gathered}$

Answer:

(a) (-3,y)

(b)

$\begin{gathered} FP=\sqrt{(x-3)^2+y^2}, \\ QP=x+3. \end{gathered}$

(c)

$FP=QP.$

(d)

$x=\frac{1}{12}y^2.$