Question

find the vertex and focus of the parabola: x² + 6x - 8y + 1 = 0 vertex = (?, ) focus = ( , )

Explanation:

Step1: Rewrite the equation

$$\begin{align*} x^{2}+6x - 8y+1&=0\\ x^{2}+6x&=8y - 1\\ x^{2}+6x + 9&=8y - 1+9\\ (x + 3)^{2}&=8y+8\\ (x + 3)^{2}&=8(y + 1) \end{align*}$$

Step2: Find the vertex

For the parabola \((x - h)^{2}=4p(y - k)\), the vertex is \((h,k)\).
Comparing \((x + 3)^{2}=8(y + 1)\) with \((x - h)^{2}=4p(y - k)\), we have \(h=-3,k = - 1\). So the vertex is \((-3,-1)\).

Step3: Find the value of \(p\)

Since \(4p = 8\), then \(p = 2\).

Step4: Find the focus

For the parabola \((x - h)^{2}=4p(y - k)\), the focus is \((h,k + p)\).
Substituting \(h=-3,k=-1,p = 2\) into \((h,k + p)\), we get \((-3,-1+2)=(-3,1)\).

Answer:

Vertex \(=(-3,-1)\)
Focus \(=(-3,1)\)