Question

find the vertex and focus of the parabola:
$y^{2}-10y + 4x + 21 = 0$
vertex = ( ?, )
focus = ( , )

Explanation:

Step1: Complete the square for \(y\) terms

$$\begin{align*} y^{2}-10y + 4x+21&=0\\ y^{2}-10y&=-4x - 21\\ y^{2}-10y + 25&=-4x-21 + 25\\ (y - 5)^{2}&=-4x + 4\\ (y - 5)^{2}&=-4(x - 1) \end{align*}$$

Step2: Use the standard form of parabola \((y - k)^{2}=4p(x - h)\)

For the parabola \((y - k)^{2}=4p(x - h)\), the vertex is \((h,k)\) and the focus is \((h + p,k)\)
Comparing \((y - 5)^{2}=-4(x - 1)\) with \((y - k)^{2}=4p(x - h)\)
We have \(h = 1,k = 5\) and \(4p=-4\Rightarrow p=-1\)

Answer:

Vertex \(=(1,5)\)
Focus \(=(0,5)\)