Math>Geometry>Parabola.....?

Question

Reduce the equation to standard from, give the coordinates of the vertex, focus, ends of the latus rectum and find the equation of the directrix. I got this wrong in my test yesterday. I'm so careless. Please as much as possible do it step by step to prevent vagueness.


36y^2+72x-132y-239=0

Answer 1

Parabola.

36y² + 72x - 132y - 239 = 0

72x = -36y² + 132y + 239
-2x = y² - (11/3)y - 239/36
-2x = [y² - (11/3)y + (11/6)²] - 239/36 - (11/6)²
-2x = (y - 11/6)² - 239/36 - 121/36
-2x = (y - 11/6)² - 10
-2x + 10 = (y - 11/6)²

-2(x - 5) = (y - 11/6)²

The vertex of the parabola is (h,k) = (5, 11/6).

The line of symmetry is a horizontal line thru the vertex. Its equation is

y = 11/6
____________

The coefficient of the x coordinate has value 4p.

4p = -2
p = -1/2
2p = -1

The directed distance from the vertex to the focus is p. The x coordinate of the focus is:

h + p = 5 - 1/2 = 9/2

The focus is (h + p, k) = (9/2, 11/6).
____________

The directrix is a vertical line perpendicular to the line of symmetry. Its x value is

h - p = 5 - (-1/2) = 11/2

The equation of the directrix is

x = 11/2
_____________

The latus rectum is a vertical line perpendicular to the line of symmetry. It runs from one side of the parabola thru the focus to the other side. Its length is 4p. It extends a distance of 2p in each direction from the focus. The y values of its endpoints are:

k ± 2p = 11/6 ± (-1) = 5/6, 17/6

The endpoints are (h + p, k ± 2p) or:

(9/2, 5/6) and (9/2, 17/6)
________

Answer 2

36y^2+72x-132y-239=0
72x=-36y^2+132y+239
divide by 4 throughout
18x=-9y^2+33y+(239/4)
18x=-(9y^2-2.3.(11/2)+(121/4)-(121/4)-(239/4))
18x=-(3y-(11/2))^2-(121/4)-(239/4)
18x=-(3y-(11/2))^2-(360/4)
18x=-(3y-(11/2))^2-90
x=-(1/18)(3y-(11/2))^2-(90/18)
x=-(1/2)(y-(11/6))^2-5
the above equation is in the standard form
vertex of the above equation would be ((11/6),-5)
i dont know how to calculate the other parameters...i hope this helps