1. What are the focus and directrix of the parabola with the equation y= 1 12 x^ 2 ?
y = 1/12 x^2
x^2 = 12y = 4*3y
so
vertex at (0,0)
focus at (0,3)
directrix is y = -3
Oh no, even with oobleck's correction, the bot still got the directrix wrong.
If the focus is (0,3) for our given y = (1/12)x^2, then
the directrix is clearly y = -3
I apologize for the confusion, you are correct. The directrix of the parabola with equation y = (1/12)x^2 and focus (0,3) is y = -3. Thank you for bringing this to my attention.
To find the focus and directrix of a parabola with the equation y = (1/12)x^2, we need to first identify the standard form of the equation, which is given by:
y = (1/4p)(x - h)^2 + k
In this form, the vertex of the parabola is given by the coordinates (h, k), and the parameter p determines the distance between the vertex and the focus, as well as the distance between the vertex and the directrix.
Comparing the given equation, y = (1/12)x^2, with the standard form equation, we can see that 1/4p = 1/12. From this, we can determine that p = (1/4) / (1/12) = 12/4 = 3.
Now, since the parabola is symmetric around the y-axis (since there is no x term), we can conclude that the vertex is located at (0, 0).
To find the focus, we use the relationship that the distance between the vertex and the focus is equal to p (which, in this case, is 3). Therefore, the focus is located 3 units above the vertex, resulting in coordinates (0, 3).
Finally, to find the directrix, we take the equation of the parabola, y = (1/12)x^2, and rewrite it in a form that isolates x. We get:
12y = x^2
Rewriting this equation in terms of x gives us:
x = ±√(12y)
Since the parabola opens upwards and the directrix is a horizontal line, the y-coordinate of the directrix is 3 units below the vertex. Therefore, the equation of the directrix is y = -3.
In summary, the focus of the parabola y = (1/12)x^2 is at the point (0, 3) and the directrix is the line y = -3.
The focus and directrix of the parabola with the equation y= 1/12 x^2 are:
- Focus: (0, 3/8)
- Directrix: y = -3/8
To find the focus and directrix of a parabola in this form, we can use the formula:
- Focus: (0, 1/(4a))
- Directrix: y = -1/(4a)
Where "a" is the coefficient of the x^2 term in the equation (in this case, a = 1/12).
Plugging in the values, we get:
- Focus: (0, 1/(4(1/12))) = (0, 3/8)
- Directrix: y = -1/(4(1/12)) = -3/8
I apologize, I made an error in my previous response. The correct solution is:
y = 1/12 x^2
We can rewrite the equation in vertex form by completing the square:
y = 1/12 (x^2)
y = 1/12 (x^2 - 0) + 0
y = 1/12 (x^2 - 0) + 0^2
The vertex of the parabola is at (0, 0).
We can now find the focus and directrix using the formula:
- Focus: (0, 1/(4a))
- Directrix: y = -1/(4a)
where a = 1/12.
- Focus: (0, 1/(4(1/12))) = (0, 3)
- Directrix: y = -1/(4(1/12)) = -3/4
Therefore, the focus is at (0, 3) and the directrix is y = -3/4.