Problem 44 - Algebra and Trigonometry

Question

In Exercises 43–48, convert each equation to standard form by completing the square on x or y. Then find the vertex, focus, and directrix of the parabola. Finally, graph the parabola. $$ x^{2}+6 x+8 y+1=0 $$

Step-by-Step Solution

Answer

The parabola's standard form is \( (x+3)^{2} = 8(y+1) \). The vertex is (-3, -1); the focus is at (-3, 1), and the directrix is the line y = -3.

Step 1: Convert to standard form

Let's first address the exercise's request to convert the equation to standard form. This requires grouping the 'x' terms to complete the square. This can be done in the following way: \( x^{2}+6 x+8 y+1=0 \) can be rewritten as \( (x^{2} + 6x) + 8y + 1 = 0 \) Now, the square can be completed by adding and subtracting \( (b/2a)^{2} \) where 'b' is the coefficient of 'x' and 'a' is the coefficient of \( x^{2} \). The equation must be kept balanced, which is why the number added must also be subtracted: \( ((x^{2} + 6x + 9) - 9) + 8y + 1 = 0 \) Therefore, \( (x+3)^{2} - 9 + 8y + 1 = 0 \) which simplifies to \( (x+3)^{2} +8y - 8 = 0 \).

Step 2: Calculate the vertex

The vertex for this form of parabola \( (x-h)^{2} = 4a(y-k) \) is given by the point (h, k). In this equation, \( h = -3 \) and \( k = -1 \). Therefore, the vertex of the parabola is (-3, -1).

Step 3: Find the focus and directrix

The value 'a' can be calculated from the standard form of the equation by dividing the 'y' coefficient by 4, \( a = 8/4 = 2 \). The focus is then (h, k + a), which is (-3, -1 + 2) = (-3, 1). The directrix is at y = k - a, which is -1 - 2 = -3. So, the directrix is given by y = -3.

Step 4: Graph the parabola

The graph of the parabola includes the vertex, focus, and the directrix. It opens upwards because the 'x' variable is squared in the equation. The vertex (-3, -1) is the bottommost point, the focus (-3, 1) is a point above the vertex inside the parabola and the line y = -3 is plotted as the directrix, a horizontal line below the vertex. A smooth curve is then drawn from the vertex, going through the focus and opening upwards.

Key Concepts

Standard Form of a Parabola

The standard form of a parabola is essential for analyzing its features like the vertex, focus, and axis of symmetry. For a vertical parabola (opening up or down), the equation is \[(x-h)^2 = 4a(y-k),\]where \((h, k)\) is the vertex.To convert an equation like \(x^2 + 6x + 8y + 1 = 0\) into standard form, we complete the square for the \(x\) terms:

• Group the \(x\) terms: \((x^2 + 6x) + 8y + 1 = 0\).
• Complete the square: Add and subtract \((3)^2\) (since \(6/2 = 3\)).
• Rewrite: \((x+3)^2 - 9 + 8y + 1 = 0\).
• Simplify: \((x+3)^2 + 8y - 8 = 0\).

This gives us the equation \((x+3)^2 = -8(y+1)\), a neat standard form for further analysis.

Vertex of a Parabola

Finding the vertex of a parabola is crucial for understanding its shape and position. In the standard form \((x-h)^2 = 4a(y-k),\)the vertex is at \((h, k).\)For the equation we converted earlier, \((x+3)^2 = -8(y+1)\):

• The terms show that \(h = -3\) and \(k = -1\).
• Thus, the vertex is at \((-3, -1)\).

The vertex acts as the parabola's lowest point if it opens upwards, or highest if it opens downwards. It helps in sketching the graph and determining the parabola’s axis of symmetry, which is the vertical line \(x = h\).

Focus and Directrix of a Parabola

The focus and directrix provide insights into the parabola's shape and orientation. From the standard form \((x-h)^2 = 4a(y-k),\)where \(a\) defines the distance:

• Focus: Located at \((h, k+a)\).
• Directrix: Described by the line \(y = k-a\).

In our example, \(a = 2\) (from \(4a = 8\)), giving:

• Focus at \((-3, 1)\).
• Directrix line at \(y = -3\).

The focus and directrix are equidistant from any point on the parabola. Bridging the vertex, they guide the curve's path.

Graphing Parabolas

Graphing a parabola accurately involves plotting its key features. Here’s a step-by-step guide:

• Identify the vertex. In this case, it's \((-3, -1)\), marking the parabola's turning point.
• Plot the focus. It’s a point inside the curve at \((-3, 1)\).
• Draw the directrix. A horizontal line at \(y = -3\), guiding the parabola's curvature.
• Shape the parabola. Since \(x\) is squared and the equation is negative, it opens upwards.

Connect these elements smoothly, ensuring the curve is symmetrical around the axis through the vertex. This visualization helps solidify understanding and provides a clear depiction of how the parabola behaves. Remember, the focus is inside the curve, showing the direction of the opening, while the directrix is outside, providing balance.