Problem 25 - Algebra for College Students

Question

For Problems \(1-30\), find the vertex, focus, and directrix of the given parabola and sketch its graph. $$ x^{2}+6 x+8 y+1=0 $$

Step-by-Step Solution

Answer

Vertex: \((-3, -1)\), Focus: \((-3, 1)\), Directrix: \(y = -3\).

Step 1: Rewrite the Equation

Start by rearranging and isolating the terms involving the same variables. Given the equation is \(x^2 + 6x + 8y + 1 = 0\), subtract 1 from both sides to separate y from x. The equation becomes \(x^2 + 6x + 8y = -1\).

Step 2: Complete the Square

To simplify, complete the square for the x-terms. Take the form \(x^2 + 6x\). Half the coefficient of \(x\) is \(3\), and \(3^2 = 9\). Add and subtract \(9\) from the equation: \((x^2 + 6x + 9) - 9 + 8y = -1\). This becomes \((x+3)^2 - 9 + 8y = -1\).

Step 3: Solve for y in Terms of x

Rearrange to express y in terms of x. The equation becomes \((x + 3)^2 = 8y - 1 + 9\), simplifying to \((x + 3)^2 = 8y + 8\) or \((x + 3)^2 = 8(y + 1)\).

Step 4: Identify Vertex, Focus, and Directrix

From \((x+3)^2 = 8(y+1)\), the vertex form \((x-h)^2 = 4p(y-k)\) gives vertex \(h = -3\), \(k = -1\). Thus, vertex \((-3, -1)\). Here, \(4p = 8\) gives \(p = 2\). The focus is \((h, k+p) = (-3, 1)\). Directrix is \(y = k-p = -3\).

Step 5: Sketch the Graph

Plot the vertex \((-3,-1)\), focus \((-3,1)\), and draw the directrix \(y = -3\). The parabola opens upwards since y is isolated and positive. Sketch the parabola with its axis of symmetry through \(x = -3\).

Key Concepts

Vertex of a Parabola

In the world of parabolas, the vertex is crucial because it serves as the starting point that determines the parabola's direction and shape. It can be thought of as the highest or lowest point of the parabola, depending on whether it opens upwards or downwards. For the parabola described by the equation

• \((x+3)^2 = 8(y+1)\), the vertex form of a parabola equation is \((x-h)^2 = 4p(y-k)\).

Here, \(h\) and \(k\) are the coordinates of the vertex. We found the vertex by comparing the given equation with the standard form, which gives us \(h = -3\) and \(k = -1\). Therefore, the vertex of the parabola is at

• \((-3, -1)\).

This indicates that from the point at

• x = -3, y = -1, the parabola begins to open.

This vertex tells us other things, too:

• The axis of symmetry runs vertically through this point.
• It helps in sketching the graph accurately.

Focus of a Parabola

The focus of a parabola is another key point that lies on its axis of symmetry. This point is crucial because every point on the parabola is equidistant from the focus and a corresponding directrix line. For our parabola, the equation

• \((x+3)^2 = 8(y+1)\) provides us the information needed to find the focus.

To identify the focus, we use the value of \(p\), which is derived from the formula \(4p = 8\), thus \(p = 2\). Knowing the vertex is

• \((-3, -1)\), and the parabola opens upward, we find the focus at \((h, k+p)\), resulting in
• \((-3, 1)\).

This focus is slightly above the vertex along the vertical axis. The focus plays a vital role:

• It helps determine how "wide" the parabola is.
• Rays emitted from the focus reflect off the parabola parallel to its axis of symmetry.

Directrix of a Parabola

The directrix of a parabola is a fixed line that is used to define the curve along with the focus. This line is crucial in understanding the symmetrical property of the parabola. For each point on the parabola, the distance to the focus is equal to the distance to the directrix. From our example

• \((x+3)^2 = 8(y+1)\), with vertex coordinates

\((-3, -1)\), and \(p = 2\), the directrix can be calculated as \(y = k-p\), which leads to

• y = -3.

The directrix isn’t part of the parabola itself, but it affects its shape and direction:

• If the parabola opens upwards, the directrix is below the vertex.
• It provides a boundary, helping to maintain that beautiful U-shape symmetry.

In summary, the directrix works hand in hand with the focus to guide the arching path of the parabola.