Problem 38 - Algebra for College Students

Question

For Problems \(31-50\), find an equation of the parabola that satisfies the given conditions. $$ \text { Focus }(-3,-1) \text {, directrix } y=7 \quad x^{2}+6 x+16 y-39=0 $$

Step-by-Step Solution

Answer

The equation of the parabola is \((x + 3)^2 = 16(y - 3)\).

Step 1: Identify Given Conditions

We have a parabola with a focus at point \((-3, -1)\) and a directrix at \(y = 7\). The given quadratic equation is \(x^2 + 6x + 16y - 39 = 0\).

Step 2: Determine Vertex and Axis of Symmetry

First, determine the vertex of the parabola. The vertex \((h, k)\) of a parabola with focus \((h, k-p)\) and directrix \(y = k+p\) implies \(-1 = k - p\) and \(7 = k + p\). Solving these, we find \(k = 3\) and \(p = 4\). Hence, the vertex is \((-3, 3)\), and the axis of symmetry is a vertical line through \(x = -3\).

Step 3: Write the Vertex Form of the Equation

The general form for a vertically oriented parabola is \((x-h)^2 = 4p(y-k)\), where \((h, k)\) is the vertex. Substituting the known values, \((x + 3)^2 = 16(y - 3)\) because \(4p = 16\) and \(p = 4\).

Step 4: Expand to Standard Form

We will expand \((x + 3)^2 = 16(y - 3)\) to check if it matches the original equation:1. First, expand \((x + 3)^2 = x^2 + 6x + 9\).2. Expand the right side: \(16(y - 3) = 16y - 48\).3. Setting the expanded forms equal gives \(x^2 + 6x + 9 = 16y - 48\).4. Rearrange to \(x^2 + 6x + 16y - 39 = 0\), which matches the given equation, confirming the equation is correct.

Key Concepts

Focus and Directrix

Every parabola has a special point called the "focus" and a line called the "directrix." These two elements play a crucial role in defining the shape and position of the parabola. The focus of the parabola is a point located at (-3, -1) in this case, and it lies inside or along the curve of the parabola. The directrix is a horizontal line at a distance from the vertex and is given by the equation \(y = 7\). This line is exactly opposite of the focus, which means if the focus is on one side of the vertex, the directrix is on the opposite side.

The defining property of a parabola is that any point on the curve is equally distant from the focus and the directrix. Mathematically, if you have a point \((x, y)\) on the parabola, the distance to the focus, \((x - (-3))^2 + (y - (-1))^2\), equals the perpendicular distance to the directrix \(|y - 7|\). This relationship helps us to establish the equation of the parabola when the focus and directrix are known.

Vertex Form of a Parabola

The vertex form of a parabola offers a convenient way of expressing the equation while making it simple to identify the vertex. For a parabola that opens vertically, the vertex form is written as

• \((x-h)^2 = 4p(y-k)\)

Here, \((h, k)\) represents the vertex of the parabola.

In this problem, the vertex \((-3, 3)\) is calculated by averaging the focus' coordinates and solving for the vertex, given the focus and directrix. Knowing \(p = 4\) allows us to write the equation in vertex form as

• \((x+3)^2 = 16(y-3)\)

This form emphasizes the transformation of the parabola in relation to the vertex, making it helpful in quickly sketching the graph and understanding its properties.

Axis of Symmetry

The axis of symmetry in a parabola is an imaginary line that perfectly cuts through the peak of the parabola, ensuring the shape on each side is a mirror image. This axis passes through the vertex and is crucial in analyzing the vertical alignment of the parabola.

For the given problem, since the parabola's vertex is at \((-3, 3)\), the axis of symmetry is a vertical line at \(x = -3\). This characteristic is particularly useful when sketching the parabola, as it helps in determining the balance and direction in which the parabola opens.

Standard Form of Quadratic Equation

The standard form provides a familiar way to write quadratic equations, typically used to identify the coefficients and apply various algebraic techniques. This form is often expressed as

• \(ax^2 + bx + c = 0\)

The equation given in the problem is partially in standard form:

• \(x^2 + 6x + 16y - 39 = 0\)

To derive this expression, we first expanded the vertex form

• \((x+3)^2 = 16(y-3)\)

which resulted in the equivalent form

• \(x^2 + 6x + 16y - 39 = 0\).

This process illustrates the equivalence between different forms of the parabola's equation, enabling the use of quadratic formulas and facilitating further problem-solving steps.