Problem 48 - College Algebra

Question

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. $$x^{2}+8 x-4 y+8=0$$

Step-by-Step Solution

Answer

In standard form, the equation of the parabola is \((x+4)^{2}=y+2\). The vertex is \((-4, -2)\), the focus is \((-4, -7/4)\), and the equation of the directrix is \(y=-9/4\).

Step 1: Convert to Standard Form

First, rearrange the equation to isolate \(y\). It will be done as follows: Add \(4y\) to both sides, then subtract 8 to obtain: \(x^{2}+8x=4y-8\). Now, the equation should be written into the standard form for a parabola that opens upwards or downwards, which is \((x-h)^{2}=4p(y-k)\). Completing the square for \(x^{2}+8x\) will complete the first side if the equation. Add \((8/2)^{2} = 16\) inside the left hand side in the equation to complete the square and do the same to the right hand side of the equation to balance. The equation becomes: \((x+4)^{2}=4y-8+16\). Simplifying right side yields \((x+4)^{2}=4y+8\). Divide all terms by 4 to get the equation in standard form: \((x+4)^{2}=y+2\).

Step 2: Find Vertex, Directrix, and Focus

The vertex of the parabola is found at point \((h, k)\) from the equation \((x-h)^{2}=4p(y-k)\). In our case, \(h = -4\) and \(k = -2\), so the vertex of the parabola is \(-4,-2\). The value of \(p\) can be derived from the equation \((x+4)^{2}=y+2\), where \(4p=1\), this means \(p=1/4\). The parabola is opened upwards since \(p>0\). The focus lies \(p\) units above the vertex. Since \(p=1/4\), we add this to the y-coordinate of the vertex to get the focus. Hence the focus of the parabola is at \(-4, -2 + 1/4 = -7/4 -1.75\). The directrix is a horizontal line \(p\) units below the vertex. Hence, the equation of the directrix will be \(y= k-p = -2 - 1/4 = -2.25\).

Step 3: Graph the Parabola

To graph, first draw an xy-coordinate plane. Plot the vertex, focus and a line indicating the directrix. Then, draw the parabola opening upwards from the vertex and passing through the focus without crossing the directrix.