Find the equation of the parabola, in standard form, whose Vertex (2, 1), ends of the latus rectum (-1, -5) and (-1, 7).
No ideas on any of these?
The length of the latus rectum 4p=12
The focus is s=(-1,1)
The vertex is A=(2,1)
The equation of the parabola is (y-k)^2= - 4a(x-h)
(y-1)^2= -12(x-2)
To find the equation of a parabola in standard form, we can start by determining its basic equation:
y = a(x - h)^2 + k
Where (h, k) represents the vertex of the parabola and "a" represents the coefficient that determines the shape and direction. In this case, the vertex (h, k) is given as (2, 1).
Now, we need to find the value of "a" in the equation. To do so, we can use the information about the ends of the latus rectum (-1, -5) and (-1, 7).
The latus rectum of a parabola is a line segment passing through the focus and perpendicular to the axis of symmetry. In this case, the vertex serves as the focus, so the latus rectum is a horizontal line segment. Therefore, we can use the coordinates of the ends of the latus rectum to find its length.
The length of the latus rectum can be calculated using the formula:
Length of latus rectum = 4a
Since we have the coordinates of the ends of the latus rectum, we can find the length:
Length of latus rectum = |y2 - y1|
= |7 - (-5)|
= 12
Substituting this value into the formula, we have:
12 = 4a
Now we can solve for "a":
a = 12 / 4
a = 3
So, we have found the value of "a" to be 3.
Substituting the vertex (h, k) = (2, 1) and the value of "a" = 3 into the equation:
y = 3(x - 2)^2 + 1
Thus, the equation of the parabola, in standard form, is y = 3(x - 2)^2 + 1.