Question

Demetrius is drafting a parabola on a coordinate plane. He plans to have the focus at  (3,1)  and the directrix at  y=−3 . What is the equation in standard form that represents the parabola?(1 point)Responsesy=−18x2−54x+38y equals negative Start Fraction 1 over 8 End Fraction x squared minus Start Fraction 5 over 4 End Fraction x plus Start Fraction 3 over 8 End Fractiony=18x2−34x+18y 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 Fractiony=−18x2−2x+1y equals negative Start Fraction 1 over 8 End Fraction x squared minus 2 x plus 1y=18x2−54x+178

Answer 1

To find the equation of the parabola, we need to use the information given about the focus and the directrix.

1. Focus: \( (3, 1) \)
2. Directrix: \( y = -3 \)

A parabola is defined as the set of points that are equidistant from the focus and the directrix. The parabola opens either upwards or downwards. Since the focus is above the directrix, the parabola opens upwards.

The vertex of a parabola that opens upwards or downwards is located halfway between the focus and the directrix.

Finding the vertex:

• The y-coordinate of the vertex is the average of the y-coordinate of the focus and the y-coordinate of the directrix: \[ y_{\text{vertex}} = \frac{1 + (-3)}{2} = \frac{-2}{2} = -1 \]
• The x-coordinate of the vertex is the same as the x-coordinate of the focus: \[ x_{\text{vertex}} = 3 \] Thus, the vertex is \( (3, -1) \).

Distance p (the distance from the vertex to the focus or the vertex to the directrix):

• The distance \( p \) can be calculated as follows: \[ p = 1 - (-1) = 2 \] The focus is 2 units above the vertex.

Equation of the parabola: Since the vertex of the parabola is at \( (h, k) = (3, -1) \) and it opens upward, the standard form of the equation of a parabola is given by: \[ (y - k) = \frac{1}{4p}(x - h)^2 \] Substituting in \( h = 3 \), \( k = -1 \), and \( p = 2 \): \[ (y + 1) = \frac{1}{8}(x - 3)^2 \]

Now, we can rewrite this in standard form by expanding it: \[ y + 1 = \frac{1}{8}(x^2 - 6x + 9) \] \[ y + 1 = \frac{1}{8}x^2 - \frac{6}{8}x + \frac{9}{8} \] \[ y = \frac{1}{8}x^2 - \frac{3}{4}x + \frac{9}{8} - 1 \] \[ y = \frac{1}{8}x^2 - \frac{3}{4}x + \frac{9}{8} - \frac{8}{8} \] \[ y = \frac{1}{8}x^2 - \frac{3}{4}x + \frac{1}{8} \]

Thus, the equation in standard form is: \[ y = \frac{1}{8}x^2 - \frac{3}{4}x + \frac{1}{8} \]

Comparing with the options provided, the matching equation is: \[ y = \frac{1}{8}x^2 - \frac{3}{4}x + \frac{1}{8} \]

The correct 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.