Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the focus and directrix of the parabola $$ x^{2}-6 x+4 y+3=0 $$

Step-by-Step Solution

Answer

Focus: (3, 1/2); Directrix: y = 5/2.

Step 1: Rearrange the Equation

Begin by reorganizing the given equation \[ x^2 - 6x + 4y + 3 = 0 \] into a form that can be more easily analyzed for the components of a parabola. Move the linear and constant terms in \( y \) to the right side:\[ x^2 - 6x = -4y - 3 \]

Step 2: Complete the Square

To convert the \( x \) terms into a perfect square, complete the square on \( x^2 - 6x \).1. Take half of the coefficient of \( x \) (which is -6), square it, and then add and subtract that value: \((-6/2)^2 = 9\).2. Add and subtract 9 within the equation:\[ x^2 - 6x + 9 - 9 = -4y - 3 \].3. This can be rewritten as:\[ (x - 3)^2 - 9 = -4y - 3 \]

Step 3: Express the Parabola in Vertex Form

Take the completed square form and move constants to the right:Add 9 to both sides to get:\[ (x - 3)^2 = -4y + 6 \]Rearrange to match the standard form of a parabola:\[ (x - 3)^2 = -4(y - \frac{3}{2}) \]

Step 4: Identify the Vertex, Focus and Directrix

The equation \((x - 3)^2 = -4(y - \frac{3}{2})\) is now in standard form \((x - h)^2 = 4p(y - k)\), where the vertex is \((h, k)\), and \(p\) determines the distance to the focus and directrix.For \((x - 3)^2 = -4(y - \frac{3}{2})\):- Vertex: \((3, \frac{3}{2})\)- Since \(4p = -4\), then \(p = -1\) (A negative indicates the parabola opens downward).The focus is located \(p\) units down from the vertex:- Focus: \((3, \frac{1}{2})\).The directrix is a line \(p\) units up from the vertex:- Directrix: \(y = \frac{5}{2}\).

Key Concepts

Vertex Form

The vertex form of a parabola is an extremely handy and expressive way to write the equation of a parabola. It highlights the vertex, which is a central feature of the parabola. The general form of a parabola in vertex form is either

• \((x - h)^2 = 4p(y - k)\) for vertical parabolas which open up or down, or
• \((y - k)^2 = 4p(x - h)\) for horizontal parabolas which open left or right.

Here,

• \((h, k)\) is the vertex of the parabola, representing the point where the parabola changes direction.
• The parameter \(p\) determines the distance between the vertex and the focus, as well as the distance from the vertex to the directrix.

For the equation \((x - 3)^2 = -4(y - \frac{3}{2})\), the vertex is \((3, \frac{3}{2})\). This form easily allows us to visualize the vertex without additional conversion, making calculations and graphing straightforward.

Focus of a Parabola

The focus of a parabola is a key point that defines the curvature and direction of the parabola. Every point on the parabola is equidistant from the focus and the directrix. In terms of the vertex form of a parabola, the focus is determined by the vertex and the value of \(p\). Specifically,

• If the parabola is opening vertically, the focus is located \(p\) units away from the vertex along the axis of symmetry.
• The formula for the focus in this situation is \((h, k + p)\) if the parabola opens upwards, and \((h, k - p)\) if it opens downwards.

In our equation \((x - 3)^2 = -4(y - \frac{3}{2})\), we found \(p = -1\), as indicated by the equation \(4p = -4\).Since the parabola opens downwards, the focus is one unit below the vertex at \((3, \frac{1}{2})\). This focus reflects the curve and direction of the parabola's downward opening.

Directrix of a Parabola

The directrix is a line that, together with the focus, characterizes a parabola's set of equidistant points. The directrix runs parallel to the parabola’s axis of symmetry and is located opposite the focus in terms of distance from the vertex.For vertical parabolas like our example,

• The directrix has a horizontal orientation.
• The directrix's equation can be found using the vertex form: it is \(y = k - p\) if the parabola opens upwards and \(y = k + p\) if it opens downwards.

In our specific case, with vertex form \((x - 3)^2 = -4(y - \frac{3}{2})\),we determined \(p = -1\).Thus, the directrix is one unit above the vertex, with the equation \(y = \frac{5}{2}\).The directrix helps guide the symmetry and shape of the parabola.

Completing the Square

Completing the square is a technique used to transform quadratic equations into a perfect square trinomial, making it simpler to understand geometrically. This method can be used to rewrite quadratic equations in vertex form, which reveals the vertex of the parabola directly.Here's how you complete the square for a given quadratic:

• Take the equation \(x^2 - 6x + 4y + 3 = 0\) and rearrange it to isolate the \(x\) terms on one side, resulting in \(x^2 - 6x = -4y - 3\).
• To complete the square on \(x^2 - 6x\), divide the coefficient of \(x\) (which is -6) by 2 and square it, yielding 9. Add and subtract this number inside the equation:\((x - 3)^2 - 9 = -4y - 3\).
• This ensures \((x - 3)^2 = -4(y - \frac{3}{2})\) reflects the structure of vertex form, highlighting the vertex of the parabola.

Completing the square is a powerful tool for translating parabolas into vertex form, providing a clearer geometric interpretation.