Question

write an equation in standard form for the parabola. a parabola with a vertex at (3,2) and a focus at (3,4). a parabola with a vertex at (4,1) and a directrix at x = 5.

Explanation:

Step1: Determine the form of the parabola

Since the \(x -\)coordinates of the vertex \((3,2)\) and focus \((3,4)\) are the same, the parabola is vertical. The standard form of a vertical parabola is \((x - h)^{2}=4p(y - k)\), where \((h,k)\) is the vertex. Here \(h = 3,k = 2\).
The distance between the vertex \((h,k)\) and focus \((h,k + p)\) is \(p\). So \(p=4 - 2=2\).
Substitute \(h = 3,k = 2,p = 2\) into \((x - h)^{2}=4p(y - k)\)
\((x - 3)^{2}=4\times2(y - 2)\)
\((x - 3)^{2}=8(y - 2)\)

Step2: For the second parabola

Since the directrix is \(x = 5\) and the vertex is \((4,1)\), the parabola is horizontal. The standard form of a horizontal parabola is \((y - k)^{2}=4p(x - h)\), where \((h,k)\) is the vertex. Here \(h = 4,k = 1\).
The distance between the vertex \((h,k)\) and directrix \(x=h - p\) is \(|p|\). Since \(h - p=5\) and \(h = 4\), then \(4-p = 5\), so \(p=-1\).
Substitute \(h = 4,k = 1,p=-1\) into \((y - k)^{2}=4p(x - h)\)
\((y - 1)^{2}=4\times(-1)(x - 4)\)
\((y - 1)^{2}=-4(x - 4)\)

Answer:

1. \((x - 3)^{2}=8(y - 2)\)
2. \((y - 1)^{2}=-4(x - 4)\)