Question

what will be equation of latus rectum of parabola x^2-2x+4y=0.

Answer 1

First, the definition of latus rectum of a parabola is the line segment perpendicular to the axis, passing through the focus of the parabola and which has its two end-points on the curve.

Refer to the illustration below:
http://imageshack.us/photo/my-images/195/1312812014.png/

The focus is at (1,-3/4) through which the red line passes. The Latus rectum is the red line segment on the interior of the curve (from -1 to 3).

To find the latus rectum, first reduce the given parabola to the form:
y=(1/4p)(x-h)^2+k
using completing the squares, and where (h,k) is the vertex. The focus is at
(h,k+p) the length of the latus rectum is 4p.

The focus is at a distance p from the vertex, interior to the curve.

For the given parabola,

x^2-2x+4y=0
reduces to
f(x)=y=-(1/4)(x^2-2x)
=-(1/4)(x-1)^2+1/4
=(1/(4(-1)))(x-1)^2+1/4
Thus the vertex is at (1,1/4) and p=-1
The focus is at (1,1/4+p)=(1,-3/4)
Since
f(-1)=-3/4, and f(3)=-3/4
we conclude that the latus rectum is between the points
(-1,-3/4) and (3,-3/4), and of length 4 (equal to 4p).

Answer 2

Given: x^2 - 2x + 2y = 0.

4y = -x^2 + 2x,
y = -(1/4)x^2 + (1/2)x,

a = -1/4,
LR = [1/a] = 4 = Latus Rectum.
4a = 4(-1/4) = -1.
1/4a = -1.
[1/4a] = 1.

h = Xv = -b/2a = (-1/2) / (-1/2) = 1.
k = Yv = -1^2/4 + (1/2)1 = 1/4.

V(1,1/4)

F(1,y)

Y = k - [1/4a] = 1/4 --1 = -3/4.

F(1,-3/4)

Eq of LR: Y = -3/4.