Problem 71 - College Algebra

Question

Write the standard form of the equation of a parabola whose points are equidistant from \(y=4\) and \((-1,0)\)

Step-by-Step Solution

Answer

The standard form of the equation of the parabola is \((x+1)^2 = 8(y-2)\)

Step 1 Identify the focus and the directrix

The focus of the parabola is given as \((-1,0)\) and the directrix is the line \(y=4\), which is a horizontal line.

Step 2 Calculate the vertex of the parabola

The vertex of a parabola is located midway between the focus and the directrix. As the focus is \((-1,0)\) and the directrix is \(y=4\), the vertex is halfway between \(y=0\) (the y-coordinate of the focus) and \(y=4\). Therefore, the y-coordinate of the vertex is \((0+4)/2 = 2\). The x-coordinate of the vertex is always the same as the x-coordinate of the focus, which is \(-1\). Thus, the vertex is \((-1,2)\).

Step 3 Use the vertex and focus to calculate the value of p

The value of \(p\) is the distance from the vertex to the focus or to the directrix. Hence, \(p = 2-0 = 2\).

Step 4 Write the equation of the parabola in standard form

Since the vertex of the parabola is \((-1,2)\), and the parabola opens downwards (since the focus is below the directrix), the equation of the parabola is written in the form \((x-h)^2 = 4p(y-k)\). Plugging in the values found gives \((x+1)^2 = 4*2(y-2)\). Simplifying this gives the final equation \((x+1)^2 = 8(y-2)\).