if equation of directrix of a parabola is 2x+3=0, axis is 2y-3=0 and length of latus rectum is 6 then what will be the focus?
Given: 2x + 3 = 0, 2y - 3 = 0, Latus
Rectum = 6.
Eq1: y = 2x + 3.
Eq2: x = 2y - 3.
In Eq1, substitute 2y - 3 for x:
y = 2(2y-3) + 3,
y = 4y - 6 + 3,
y - 4y = -3,
-3y = -3,
y = 1.
In Eq2, substitute 1 for y:
x = 2*1 - 3 = -1.
D(-1,1).
1/a = 6,
a = 1/6.
4a = 4/6 = 2/3.
1/4a = 3/2.
F(-1,y)
V(-1,5/2)
D(-1,1)
k = 1 + 1/4a = 1 + 3/2 = 5/2 = 2 1/2.
y = k + 1/4a = 5/2 + 3/2 = 8/2 = 4.
To find the focus of a parabola given the equation of the directrix, the equation of the axis, and the length of the latus rectum, follow these steps:
Step 1: Rewrite the equation of the directrix in the standard form.
2x + 3 = 0
Subtract 3 from both sides:
2x = -3
Divide by 2:
x = -3/2
Step 2: Rewrite the equation of the axis in the standard form.
2y - 3 = 0
Add 3 to both sides:
2y = 3
Divide by 2:
y = 3/2
Step 3: Determine the distance between the directrix and the axis.
The distance between the directrix and the axis is the perpendicular distance between them. It can be found by taking the absolute value of the coefficient of x (or y, depending on the orientation of the parabola):
Distance = |coefficient of x (or y)| = |2| = 2
Step 4: Determine the coordinates of the focus.
The focus lies on the axis of symmetry, which is parallel to the directrix and equidistant from it. Since the distance between the directrix and the axis is 2, the focus will also be 2 units away from the axis of symmetry.
For a parabola with equation y^2 = 4ax, the distance between the focus and the vertex is a. Since the focus is 2 units away from the axis of symmetry, a = 2.
Now we can find the x-coordinate of the focus by subtracting a (2) from the x-coordinate of the vertex (which is -3/2):
x-coordinate of focus = -3/2 - 2 = -7/2
The y-coordinate of the focus is the same as the y-coordinate of the vertex (3/2).
Therefore, the coordinates of the focus are (-7/2, 3/2).