Question

find the vertex and focus of the parabola: $x^2 + 6x + 16y + 121 = 0$ vertex = (?, ) focus = (, )

Explanation:

Step1: Complete the square for x-terms

Given the equation \(x^{2}+6x + 16y+121 = 0\). First, group the \(x\)-terms: \((x^{2}+6x)+16y + 121=0\). To complete the square for \(x^{2}+6x\), we take half of 6 (which is 3), square it (which is 9), so we add and subtract 9: \((x^{2}+6x + 9-9)+16y+121 = 0\). This can be rewritten as \((x + 3)^{2}-9+16y+121 = 0\).

Step2: Simplify the equation

Simplify the constants: \((x + 3)^{2}+16y+112 = 0\). Then, isolate the squared term: \((x + 3)^{2}=-16y - 112\). Factor out -16 from the right - hand side: \((x + 3)^{2}=-16(y + 7)\).

The standard form of a parabola that opens up or down is \((x - h)^{2}=4p(y - k)\), where \((h,k)\) is the vertex and \((h,k + p)\) is the focus. Comparing \((x + 3)^{2}=-16(y + 7)\) with \((x - h)^{2}=4p(y - k)\), we have \(h=-3\), \(k=-7\), and \(4p=-16\), so \(p=-4\).

Step3: Find the vertex and focus

• Vertex: The vertex \((h,k)\) is \((-3,-7)\) since \(h=-3\) and \(k = - 7\).
• Focus: The focus is \((h,k + p)\). Substituting \(h=-3\), \(k=-7\), and \(p=-4\), we get the focus as \((-3,-7-4)=(-3,-11)\).

Answer:

Vertex \(=(-3,-7)\)
Focus \(=(-3,-11)\)