Question

write an equation of a parabola that opens up, with the given focus and directrix. focus: (0, 2) directrix: y = -1 y = < (x - < )² + <

Explanation:

Step1: Find the vertex

The vertex of a parabola is the mid - point between the focus \((0,2)\) and the directrix \(y = - 1\).
The \(y\) - coordinate of the vertex \(k=\frac{2+( - 1)}{2}=\frac{1}{2}\), and the \(x\) - coordinate of the vertex \(h = 0\) (since the parabola is symmetric about the line \(x = h\) which is \(x = 0\) here as the focus and directrix are symmetric about \(x = 0\)). So the vertex is \((h,k)=(0,\frac{1}{2})\).

Step2: Find the value of \(p\)

The distance between the focus \((0,2)\) and the vertex \((0,\frac{1}{2})\) is \(p\). Using the formula \(p=y_{focus}-y_{vertex}\), we have \(p = 2-\frac{1}{2}=\frac{3}{2}\).

Step3: Write the equation of the parabola

The standard form of a parabola that opens up is \(y=\frac{1}{4p}(x - h)^{2}+k\).
Substituting \(h = 0\), \(k=\frac{1}{2}\), and \(p=\frac{3}{2}\) into the formula:

$$
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$$

Answer:

\(y=\frac{1}{6}x^{2}+\frac{1}{2}\)