Problem 90 - A Graphical Approach to Precalculus with Limits

Question

Find an equation of a parabola that satisfies the given conditions. Focus \((1,2) ;\) directrix \(y=4\)

Step-by-Step Solution

Answer

The equation of the parabola is \((x-1)^2 = 4(y-3)\).

Step 1: Understand Parabola Definition

A parabola is defined as the set of points that are equidistant from a point called the focus and a line called the directrix.

Step 2: Recall the General Equation for Parabolas

For a vertical parabola where the vertex is at point \((h, k)\), the equation is \((x-h)^2 = 4p(y-k)\). 'p' is the distance between the vertex and the focus, and the directrix is \(y=k-p\).

Step 3: Calculate Vertex Position

Given the focus \((1, 2)\) and directrix \(y=4\), remember the vertex is halfway between the focus and directrix vertically. Calculate: \(k = \frac{2+4}{2} = 3\). Thus, vertex is at \((1, 3)\).

Step 4: Calculate 'p' Value

The distance \(p\) is the difference between the y-coordinate of the vertex and either the focus or the directrix. Here, \(p = 3 - 2 = 1\), since the focus is below the vertex.

Step 5: Write the Final Equation

Substitute \(h=1\), \(k=3\), and \(p=1\) into the parabola equation: \((x-1)^2 = 4(1)(y-3)\), which simplifies to \((x-1)^2 = 4(y-3)\).

Key Concepts

Focus and Directrix

One of the fundamental concepts of parabolas is understanding the roles of the focus and directrix. These two elements define the geometric property of a parabola.
The focus is a fixed point. For the parabola in our exercise, it is located at \((1, 2)\). The directrix, on the other hand, is a fixed line. Here, it is given by the equation \(y=4\).
Every point on the parabola is equidistant from the focus and the directrix. This relationship is key to understanding the shape and position of the parabola. To visualize this, imagine that for every point \((x, y)\) on the parabola, the distance to the focus \((1, 2)\) is exactly the same as the distance to the line \(y=4\). This balance results in the distinct 'U' shape of the parabola.

Vertex Calculation

The vertex of a parabola represents its highest or lowest point, depending on its orientation. In a vertical parabola, which opens up or down, the vertex is midway between the focus and directrix.
For our problem, we have the focus at \((1, 2)\) and the directrix \(y=4\). The vertex, being halfway between these points vertically, is calculated as:

• Sum the y-coordinates of the focus and directrix: \(2 + 4 = 6\)
• Divide by 2 to find the midpoint: \(\frac{6}{2} = 3\)

The vertex has the same x-coordinate as the focus, which is 1, giving us the vertex \((1, 3)\).
The vertex plays a crucial role in forming the equation of the parabola, as it serves as the point \((h, k)\) in the formula.

Distance 'p'

The letter \(p\) in the parabola equation is crucial. It represents the directed distance between the vertex and the focus or from the vertex to the directrix.
In this exercise, the vertex is at \((1, 3)\) and the focus at \((1, 2)\). To find \(p\), calculate the vertical difference between the vertex and the focus:
\[p = 3 - 2 = 1\]
This distance shows that the vertex is 1 unit away from the focus. Since the focus is below the vertex, this implies \(p\) is positive, confirming the parabola opens upwards.
A positive \(p\) for a vertical parabola indicates it opens upwards, while a negative \(p\) would imply it opens downwards.

Vertical Parabola

A vertical parabola either opens upwards or downwards. This orientation is determined by the value of \(p\).
If \(p\) is positive, the parabola opens upward. If \(p\) is negative, it opens downward.
Using the formula \((x-h)^2 = 4p(y-k)\), where \((h, k)\) is the vertex, we observe:

• Our calculations give \(h = 1\), \(k = 3\), and \(p = 1\).

Substituting these into the equation, we get:
\[(x-1)^2 = 4(1)(y-3)\]
which simplifies to:
\[(x-1)^2 = 4(y-3)\]
This equation represents a vertical parabola with its vertex at \((1, 3)\) opening upwards. Understanding this format helps in identifying the parabola's direction and position on the Cartesian plane.