Question

Find the standard equation of the parabola with a focus at (0,2) and directrix at y=4; also find the latus rectum.

Answer 1

Insert points (0,2) and (x,4) into this formula

\[(x - x_1)^2 + (y - y_1)^2 = (x - x_2)^2 + (y - y_2)^2\]

To get:

\[(x - 0)^2 + (y - 2)^2 = (x - x)^2 + (y - 4)^2\]

From there, simplify, expand and isolate y

Answer 2

first find the vertex which is the mid-point of (0,2) and (0,4) here
and then shift the orifin to the vertex and refer to standard forms..

Answer 3

here the vertex of the required parabola is at (0,3)
and the distance between vertex and focus is 1 unit... (distance between (0,2) and (0,3))

Answer 4

let x-0=X and y-3=Y ==>> x=X and y=Y+3
hence the standard with reference to new axes is given by X^2=-4*1*Y
which when translated to old axes
may be written as x^2=-4(y-3)...

Answer 5

latus rectum is line perpendicular to the axis and passing through the focus
thus in this case it is y=2

Answer 6

@matricked Thank you for your in-depth explanation! That clarified a lot, but now my only confusion is how a = 1. I know that when the vertex is at (0,0) and a>0, the focus is (0,-a) - which in this case is 2. What am I misunderstanding?

Answer 7

a is the focal length whic is the distance beteen th vertex and the focus..
here distance between (0,2) and (0,3)

Answer 8

@matricked That makes total sense. I wish my textbook would define > imply

Answer 9

it must be there ..and i know it is there
its we don't see things properly..

Answer 10

To finish what I started:

\(x^2 + y^2 - 4y + 4 = y^2 - 8y + 16\)

\(x^2 + y^2 - y^2 + 4 - 16 = 4y - 8y\)

\(x^2 - 12 = -4y\)

\(\dfrac{x^2 - 12}{-4} = y\)

\(-\dfrac{x^2}{4} + 3 = y\)