find the equation of parabola with latus rectum joinin.g (2, 5), (2, -3).
find the equation of parabola with bertex in tje line y=6, axis parallel to y, latus rectum 6 and passing through (2, 8)
Recall that the parabola y^2 = 4px has
vertex at (0,0)
focus at (p,0)
directrix at x = -p
latus rectum of length 4p
You say the latus rectum has length=6, so I don't know what those two points are all about. They are clearly not the ends of the latus rectum. Especially since that line is vertical, yet you say the axis is vertical. I'll work a solution assuming a horizontal axis. You can revise it as you see fit.
So, let's say that the vertex is on the line y=6. That means we have
(y-6)^2 = 3/2 (x-h)
Since (2,8) is on the curve,
3/2 (2-h) = (8-6)^2
3 - 3/2 h = 4
h = 2/3
So, (y-6)^2 = 3/2 (x - 2/3)
#1.
Since the latus rectum is a vertical line, the parabola has a horizontal axis of
symmetry and takes the general form y^2 = 4ax, where 4a is the length of the
latus rectum.
4a = √((2-2)^2 + (5+3)^2) = 8
a = 2
Also we know that the focus is (2,1) , the midpoint of the latus rectum
so we know the vertex must be (0,1) since the the distance between the focus
and the vertex is a.
equation of parabola:
(y-1)^2 = 8x <------ expand it if need be
check:
is the point (2,5) on this?
LS = (y-1)^2 = 16
RS = 8(2) = 16, YES!
#2.
The latus rectum = 6, so 4a = 6
a = 3/2
Let the vertex be (p,6) and the equation is
(y-6) = 6(x-p)^2 **
but (2,8) lies on this, so
(8-6) = 6(2-p)^2
1/3 = (2-p)^2
2-p = ±1/√3 = ± √3/3
p = 2+√3/3 or p = 2-√3/3
= (6 + √3)/3 or (6 - √3)/3
sub p into ** and you will have 2 parabolas
check my arithmetic, I was expecting a "nicer" answer.
To find the equation of a parabola, we need to know its focus and directrix. However, in both of the given cases, we are provided with the latus rectum and a point on the parabola.
Case 1: Latus Rectum given (joining (2,5) and (2,-3))
Latus Rectum is the line segment passing through the focus and perpendicular to the axis of the parabola.
Step 1: Find the midpoint of the line segment joining the two given points:
Midpoint coordinates = (2, (5+(-3))/2) = (2,1)
Step 2: The midpoint lies on the axis of the parabola. Therefore, the axis of the parabola is the vertical line x = 2.
Step 3: The distance between the midpoint and any of the given points is the latus rectum.
Distance between (2,1) and (2,5) = 4 units
Therefore, the latus rectum = 4
Step 4: The focus lies on the axis, equidistant from the vertex and directrix. The distance between the midpoint and the directrix is half of the latus rectum.
Distance between (2,1) and the directrix = 4/2 = 2 units
Step 5: Since the directrix is perpendicular to the axis, the equation of the directrix is the line parallel to the x-axis, which passes through the point (2,1+2) = (2,3)
Step 6: Since the focus lies on the axis, its x-coordinate is the same as the vertex (2,1). The y-coordinate of the focus can be found by subtracting the distance between the focus and directrix from the y-coordinate of the vertex.
Y-coordinate of the focus = 1 - 2 = -1
Therefore, the focus is at the point (2, -1).
Step 7: Knowing the focus and the directrix, we can find the equation of the parabola using the formula:
(y - y1)^2 = 4a(x - x1)
where (x1, y1) is the vertex and "a" is the distance between the vertex and the focus.
Substituting the values:
(y - 1)^2 = 4(-1)(x - 2)
(y - 1)^2 = -4(x - 2)
y^2 - 2y + 1 = -4x + 8
Therefore, the equation of the parabola is y^2 - 2y + 1 = -4x + 8.
Case 2: Vertex given (in the line y = 6, axis parallel to y, latus rectum = 6, passes through (2, 8))
Since the axis is parallel to the y-axis and the latus rectum of the parabola is given, the focus will lie on the axis, equidistant from the vertex and directrix.
Step 1: The vertex of the parabola is given as (x, y) = (2, 6).
Step 2: The distance between the vertex and the focus is half the latus rectum.
The distance between the vertex and the focus = latus rectum / 2 = 6 / 2 = 3 units.
Step 3: The focus lies on the axis and has the same x-coordinate as the vertex. Therefore, the focus is at the point (2, 6+3) = (2, 9).
Step 4: The directrix is a horizontal line parallel to the x-axis and is at a distance of 3 units from the vertex in the opposite direction of the focus. Therefore, the equation of the directrix is y = 6 - 3 = 3.
Step 5: Using the formula y = a(x - h)^2 + k, where (h, k) is the vertex, we can substitute the vertex coordinates (2, 6) and the focus coordinates (2, 9) to find the value of "a".
6 = a(2 - 2)^2 + 6
a(0) = 0
Therefore, a = 0.
Step 6: Substituting the value of "a" in the equation y = a(x - h)^2 + k, we get:
y = 0(x - 2)^2 + 6
y = 6
Therefore, the equation of the parabola is y = 6.