Find the focus and directrix of the parabola y=3x^2-12x+17
To find the focus and directrix of the parabola y = 3x^2 - 12x + 17, we first rewrite the equation in standard form:
y = 3x^2 - 12x + 17
y = 3(x^2 - 4x) + 17
y = 3(x^2 - 4x + 4) - 3(4) + 17
y = 3(x - 2)^2 + 5
Now the parabola is in the form y = a(x - h)^2 + k, where (h,k) is the vertex of the parabola. In this case, the vertex is (2,5).
The focus of the parabola is a point on the axis of symmetry which is located inside the parabola at a distance of |1/(4a)| units from the vertex, where a is the coefficient of the squared term.
In this case, a = 3, so the distance from the vertex to the focus is |1/(4*3)| = 1/12 units.
Since the parabola opens upwards, the focus will be 1/12 units higher than the vertex, so the focus will be at (2, 5 + 1/12) = (2, 61/12).
The directrix of the parabola is a horizontal line located outside the parabola at a distance of |1/(4a)| units from the vertex in the opposite direction.
In this case, the directrix is below the vertex, so it will be 1/12 units below the vertex, meaning it is at y = 5 - 1/12 = 60/12 - 1/12 = 59/12.
Therefore, the focus of the parabola is at (2, 61/12) and the directrix is at y = 59/12.