Problem 23
Question
Find an equation of the parabola that satisfies the given conditions. Focus \(F(6,4), \quad\) directrix \(y=-2\)
Step-by-Step Solution
Verified Answer
The equation of the parabola is \( (x-6)^2 = 12(y-1) \).
1Step 1: Understanding the Parabola Definition
A parabola is the set of all points equidistant from a point called the focus and a line called the directrix. Here, the focus is at \( (6,4) \) and the directrix is \( y = -2 \).
2Step 2: Using the Vertex Form of the Equation
For a vertical parabola, the equation can be written in the form \( (x - h)^2 = 4p(y - k) \) where \( (h, k) \) is the vertex and \( p \) is the distance from the vertex to the focus.
3Step 3: Finding the Vertex
The vertex lies midway between the focus and the directrix. The vertical coordinate of the vertex \( k \) is the average of the y-coordinate of the focus and y-value of the directrix: \( k = \frac{4 + (-2)}{2} = 1 \). So the vertex is at \( (6, 1) \).
4Step 4: Determining the Value of \( p \)
The distance \( p \) is the absolute difference between the y-coordinate of the vertex and the focus or the directrix. Since the vertex is at \( y = 1 \), \( p = |4 - 1| = 3 \).
5Step 5: Writing the Equation
Substitute \( h = 6 \), \( k = 1 \), and \( p = 3 \) into the vertex form equation: \( (x-6)^2 = 4 \times 3 (y-1) \), which simplifies to \((x-6)^2 = 12(y-1) \).
Key Concepts
Focus and DirectrixVertex FormEquations of ParabolasVertex of a Parabola
Focus and Directrix
In parabolas, understanding the relationship between focus and directrix is fundamental. A parabola is defined as the set of all points that are equidistant from a point known as the focus, and a line called the directrix. This means that if you take any point on the curve of the parabola, its distance to the focus is the same as its perpendicular distance to the directrix. This unique property helps us derive the equation of any parabola if the focus and directrix are known. For example, given a focus at \((6,4)\) and a directrix \(y = -2\), we can define the vertical position and shape of the parabola.
Vertex Form
The vertex form of a parabola's equation is very useful for understanding its geometry instantly. For vertical parabolas (opening upwards or downwards), the vertex form is expressed as: \[ (x - h)^2 = 4p(y - k) \] In this expression, \((h,k)\) represents the vertex of the parabola, and \(p\) denotes the distance from the vertex to the focus, which also equals the distance from the vertex to the directrix. This form is particularly convenient because it reveals the location of the vertex and the "width" of the parabola along the \(y\)-axis at a glance. By plugging in specific values for \(h, k,\) and \(p\), the full equation can describe specific shapes.
Equations of Parabolas
Parabolas can be expressed in multiple forms, but the vertex form is often very intuitive for quickly understanding key properties. Once we have values for the vertex's coordinates \((h, k)\) and the parameter \(p\), representing the focal length, these can be substituted into the vertex form equation \[ (x - h)^2 = 4p(y - k) \] to get the full equation. For practical purposes, let's explore how these values integrate into the equation:
- If \(h = 6\) and \(k = 1\), we insert these into the equation.
- With \(p = 3\), being the distance to either the focus or directrix, the equation becomes \((x-6)^2 = 12(y-1)\).
Vertex of a Parabola
The vertex of a parabola provides crucial information about the curve's position and symmetry. It is located halfway between the focus and the directrix. If you're given the focus \((6,4)\) and directrix \(y = -2\), you can calculate the vertex's \(y\)-coordinate by averaging the \(y\)-position of the focus and the \(y\)-value of the directrix: \(k = \frac{4+(-2)}{2} = 1\). Thus, the vertex is positioned at \((6, 1)\). Since parabolas are symmetric, the vertex not only marks the turning point of the curve but also acts as the axis of symmetry.
Other exercises in this chapter
Problem 23
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Find an equation for the hyperbola that has its center at the origin and satisfies the given conditions. Foci \(F(0, \pm 3)\) vertices \(V(0, \pm 2)\)
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