The given equation (x^2 = 12y) represents a parabola that opens upwards. The length of the focal width (the distance between the two points on the parabola where it intersects a line parallel to the directrix) is equal to (4p), where (p) is the distance from the vertex to the focus. For this parabola, (4p = 12), so the focal width is (12). Therefore, (12) is the length of the focal width.
12y + 5 = 53 12y = 48 y = 4
12y+2=146 12y=146-2 12y=144 144/12 12x12=144 y=12
12y = 132y = 132/12y = 11
7x - 12y = any other number.
12y - 4 = 7y + 165y = 20y = 4
12y + 5 = 53 12y = 48 y = 4
12y+2=146 12y=146-2 12y=144 144/12 12x12=144 y=12
-6x + 12y = -102 -6x = -12y - 102 x = 2y + 17
12y = 132y = 132/12y = 11
y = 11
Answer:12y=132y=132/12=11
7x - 12y = any other number.
5x=11y
12y - 4 = 7y + 165y = 20y = 4
The answer is all the points on the line defined by 25x - 12y = 1
6x - 12y = 24 Algebraically rearrange -12y = -6x + 24 Multiply both sides by '-1' Hence 12y = 6x - 24 Factor out '6' on both sides. 2y = x -4 Divide both sides by '2' y = (1/2)x - 2 Hence the gradient is ' 1/2' and the y-intersect is ' - 2' .
12y+3=9y-15 3y=-18 y=-6