(I posted this yesterday, if it looks familar, but no one was on who could help me). I am supposed to graph y+3=-1/12(x-1)^2. How do I find the equation of the directrix, and the coordinates of the vertex and the focus?
It is a downward-aimed parabola with vertex (highest value) at x = 1, y = -3.
The (1/12) coefficient in front of the (x-1)^2 term tells you about the shape of the parabola. The minus sign tells you tahat it points downward.
If you review your analytic geometry text section about parabolas, you will find that the coefficient (1/12 in this case) is called 1/(4p), where p is the distance from the vertex to the focal point and (in the other direction) to the directrix line. In your case 1/4p = 1/12, so 4p = 12 and p = 3.
The focus is therefore at y=-6 (three units below the vertex) @ x = 1, and the directrix is (y = 0), the x-axis.
y+3=-(1/12)(x-1)^2 I assume
At x = 1, the right hand side is zero.
a little to the right of x = 1, the right hand side is negative
a little to the left of x = 1, the right hand side is negative by the same amount.
In other words, the line is symmetric about the vertical line x = 1 and opens down.
SO
The vertex lies on x = 1
When x = 1, y = -3
SO
vertex at (1,-3)
Now you want the focus and directrix
The focus must also lie on x = 1
The form is
(x-h)^2 = 4 a (y-k)
here
(x-1)^2 = -12 (y+3)
so 4 a = -12
a = -3
vertex to focus = a = -3
so focus at -3-3 = -6 so (1,-6)
directrix is horizontal line a units above vertex so it is the line y = 0
To find the equation of the directrix, as well as the coordinates of the vertex and the focus of the given equation, let's work step by step:
Step 1: Identify the standard form of the equation
The equation you provided, y + 3 = -(1/12)(x - 1)^2, is in vertex form: y = a(x - h)^2 + k. Here, (h, k) represents the vertex of the parabola.
Step 2: Identify the vertex
Comparing the given equation with the standard form, we can see that h = 1 and k = -3. Therefore, the vertex of the parabola is at (1, -3).
Step 3: Find the value of "a"
From the given equation, we can see that the coefficient 'a' is -(1/12). The value of 'a' determines the shape and direction of the parabola.
Step 4: Calculate the directrix
The directrix of a parabola is a straight line perpendicular to the axis of symmetry. Its equation is given by:
y = k - a
Substituting the values of k and a, we get:
y = -3 - (-(1/12))
Simplifying further:
y = -3 + 1/12
y = -35/12
Therefore, the equation of the directrix is y = -35/12.
Step 5: Calculate the focus
The distance between the vertex and the focus is denoted by 'p' and is given by p = 1 / (4a). In this case, a = -(1/12), so:
p = 1 / (4 * -(1/12))
Simplifying further:
p = -3
Since the parabola opens downwards, the focus will be located above the vertex. So, to get the coordinates of the focus, we move p units above the vertex.
For the given equation, the coordinates of the focus are (1, -3 - 3) = (1, -6).
So, to summarize:
- The vertex is at (1, -3).
- The equation of the directrix is y = -35/12.
- The focus is at (1, -6).