Question

write an equation of a parabola that opens down, with the given focus and directrix.
focus: (0, -3) directrix: y = 1
$y = square (x - square )^{2} + square$

Explanation:

Step1: Find the vertex

The vertex \((h,k)\) is the mid - point between the focus \((0,-3)\) and the directrix \(y = 1\).
The formula for the \(y\) - coordinate of the vertex \(k=\frac{-3 + 1}{2}=-1\), and \(h = 0\) (since the \(x\) - coordinate of the focus is \(0\) and the parabola is symmetric about the line \(x = h\)). So the vertex is \((0,-1)\).

Step2: Find the value of \(p\)

The distance \(p\) from the vertex to the focus (or from the vertex to the directrix). Since the focus is \((0,-3)\) and the vertex is \((0,-1)\), then \(p=-3-(-1)=-2\) (negative because the parabola opens down).

Step3: Use the standard form of the parabola equation

The standard form of a parabola that opens up or down is \(y=a(x - h)^{2}+k\), and \(a=\frac{1}{4p}\). Substituting \(h = 0,k=-1,p=-2\), we get \(a=\frac{1}{4\times(-2)}=-\frac{1}{8}\).
The equation is \(y=-\frac{1}{8}(x - 0)^{2}-1\)

Answer:

\(y=-\frac{1}{8}x^{2}-1\)