To find the equation of the parabola with the focus at (3, 1) and the directrix at \( y = -3 \), we will use the definition of a parabola: it is the set of points equidistant from the focus and the directrix.
First, the vertex of the parabola lies midway between the focus and the directrix.
- The focus is at \( (3, 1) \).
- The directrix line is \( y = -3 \).
The y-coordinate of the vertex can be calculated as follows: \[ \text{Vertex y-coordinate} = \frac{1 + (-3)}{2} = \frac{-2}{2} = -1 \] The x-coordinate of the vertex is the same as that of the focus: \[ \text{Vertex} = (3, -1) \]
The distance \( p \) from the vertex to the focus (or to the directrix) is given by: \[ p = 1 - (-1) = 2 \] This means \( p = 2 \) and since the focus is above the directrix, the parabola opens upwards.
The standard form of the equation of a parabola that opens upwards (or downwards) is given by: \[ (y - k) = \frac{1}{4p}(x - h)^2 \] where \( (h, k) \) is the vertex of the parabola and \( p \) is the distance from the vertex to the focus.
Substituting in the values, we have:
- \( h = 3 \)
- \( k = -1 \)
- \( p = 2 \)
Thus the equation becomes: \[ (y + 1) = \frac{1}{4 \cdot 2}(x - 3)^2 \ (y + 1) = \frac{1}{8}(x - 3)^2 \]
Now, we can expand this to put it in standard form: \[ y + 1 = \frac{1}{8}(x^2 - 6x + 9) \ y + 1 = \frac{1}{8}x^2 - \frac{6}{8}x + \frac{9}{8} \] Subtracting 1 from both sides: \[ y = \frac{1}{8}x^2 - \frac{6}{8}x + \frac{9}{8} - 1 \ y = \frac{1}{8}x^2 - \frac{3}{4}x + \frac{1}{8} \]
Thus, the equation of the parabola in standard form is: \[ y = \frac{1}{8}x^2 - \frac{3}{4}x + \frac{1}{8} \]
Looking at the response options given, the match is: \[ y = \frac{1}{8}x^2 - \frac{3}{4}x + \frac{1}{8} \]
Thus, the answer is: y equals Start Fraction 1 over 8 End Fraction x squared minus Start Fraction 3 over 4 End Fraction x plus Start Fraction 1 over 8 End Fraction.