Question

Find the focus of theparabola:y2 - 10y + 4x + 21 = 0 vertex (1,5)

Answer 1

We need to express the quadratic function as follows:

`(y-k)^2=4p(x-h)`

First, we need to complete the square, as follows:

`\begin{gathered} y^2-10y+4x+21=0 \\ \text{adding and subtracting 25:} \\ y^2-10y+25-25+4x+21=0 \\ (y-5)^2-25+4x+21=0 \\ (y-5)^2+4x-4=0 \end{gathered}`

Isolating the quadratic term:

`\begin{gathered} (y-5)^2=-4x+4 \\ \text{Taking -4 as GCF:} \\ (y-5)^2=-4(x-1) \end{gathered}`

This expression has the same form as the general expression of the beginning with:

k = 5

p = -1

h = 1

The focus is placed at the point (h+p, k). In this case: (1 + (-1), 5) = (0, 5)