Question

b. focus $(1,-3)$, directrix $y=-7$
equation: _

Explanation:

Step1: Find the vertex

The vertex \((h,k)\) is the mid - point between the focus \((x_f,y_f)=(1,-3)\) and the directrix \(y = y_d=-7\).
The \(x\) - coordinate of the vertex \(h = 1\) (since the focus and the directrix are symmetric about the line \(x = h\) for a parabola of the form \((x - h)^2=4p(y - k)\)).
The \(y\) - coordinate of the vertex \(k=\frac{-3+( - 7)}{2}=\frac{-3 - 7}{2}=\frac{-10}{2}=-5\). So the vertex is \((h,k)=(1,-5)\).

Step2: Find the value of \(p\)

The distance between the vertex \((h,k)=(1,-5)\) and the focus \((h,k + p)=(1,-3)\) (or the distance between the vertex and the directrix \(y=k - p\)).
Using the formula \(k + p=-3\), substitute \(k=-5\) into it. Then \(-5 + p=-3\), so \(p = 2\).

Step3: Write the equation of the parabola

The standard form of a parabola with vertex \((h,k)\) and vertical axis of symmetry is \((x - h)^2=4p(y - k)\).
Substitute \(h = 1,k=-5,p = 2\) into the formula: \((x - 1)^2=4\times2\times(y+5)\).

Answer:

\((x - 1)^2=8(y + 5)\)