Find the Vertex, Focus, and Directrix of the parabola:
x^2 -12x-48y+276=0
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identify the focus,directrix,and axis of symmetry of f(x)= 1/16 x^2
17
To find the vertex, focus, and directrix of the parabola in the given equation, we need to first put the equation in the standard form:
y = a(x - h)^2 + k
where (h, k) is the vertex and 'a' determines the shape and orientation of the parabola.
Let's start by rearranging the equation:
(x^2 -12x) - 48y + 276 = 0
Rearrange the terms:
x^2 - 12x = 48y - 276
Now, complete the square for the x terms:
(x^2 - 12x + 36) = 48y - 276 + 36
(x - 6)^2 = 48y - 240
Now, add 240 to both sides:
(x - 6)^2 + 240 = 48y
Rearrange to isolate y:
y = (1/48)(x - 6)^2 + 5
Now we can see that the parabola is in the standard form.
Comparing this equation with the standard form equation y = a(x - h)^2 + k, we can determine the vertex, focus, and directrix.
- The vertex is at (h, k), which is (6, 5) in this case.
- The value of 'a' in this equation is 1/48, which tells us the parabola is narrow.
- The focus is located at (h, k + 1/(4a)). Therefore, the focus is at (6, 5 + 48) = (6, 53).
- The directrix is given by the equation y = k - 1/(4a). Plugging in the values, we get y = 5 - 48 = -43.
Therefore, the vertex is (6, 5), the focus is (6, 53), and the directrix is the horizontal line y = -43.
I hope this helps! Let me know if you have any further questions.