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Chapter 7: Problem 47

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. $$x^{2}+6 x-4 y+1=0$$

Short Answer

Expert verified
The parabola's standard form equation is \((x+3)^2 = 4*(y+2)\), with a vertex at (-3, -2), a focus at (-3, -1) and a directrix given by the equation \(y = -3\). This parabola opens upwards.

Step by step solution

01

Rearrange and Complete the Square

The given equation is \(x^{2}+6x-4y+1=0\). Isolate the \(x\) terms on one side and the rest on the other side to get \(x^{2} + 6x = 4y - 1\). Now, we complete the square by finding (\(b/2)^2\), where \(b\) is the coefficient of \(x\), which in this case is 6. Thus, the square to complete is \((6/2)^2 = 9\). So, we add and subtract 9 on the \(x\)-side of the equation to get \(x^{2} + 6x + 9 = 4y - 1 + 9, x^{2} + 6x + 9 = 4y + 8\). The left side can now be rewritten as \((x+3)^2 = 4y + 8\).
02

Convert to Standard Form

Rearrange the last equation to get it into the standard form: \((x-h)^2 = 4p(y-k)\) where \((h,k)\) are the coordinates of the vertex and \(p\) is the distance from the vertex to the focus. So we get \((x + 3)^2 = 4*(y +2)\).
03

Find the Vertex, Focus, and Directrix

From the standard form \((x+3)^2 = 4*(y+2)\), we can see that the vertex of the parabola is at \((-3,-2)\). Since \(4p=4\) (the coefficient of \(y\) on the right-hand side), \(p = 1\). The focus lies \(p\) units above the vertex because the parabola opens upward. Thus, the focus is \((-3, -2+1) = (-3,-1)\). Also, the directrix is \(p\) units below the vertex and thus, its equation is \(y = -2 - 1 = -3\).
04

Sketch the Parabola

Plot the vertex, focus and draw the directrix. The parabola should be shaped around the focus, away from the directrix. It will open upwards, with vertex at (-3,-2), focus at (-3,-1) and directrix at \(y=-3\).

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