Question

What is the focus of the parabola `x-4y^2+16y-19=0`?

Answer 1

The coordinates of focus of the given parabola are `(49/16,2).`

Explanation:

`x-4y^2+16y-19=0`
`implies 4y^2-16y+16=x-3`
`implies y^2-4y+4=x/4-3/4`
`implies (y-2)^2=4*1/16(x-3)`
This is a parabola along x-axis.
The general equation of a parabola along x-axis is `(y-k)^2=4a(x-h)`,
where `(h,k)` are coordinates of vertex and `a` is the distance from vertex to the focus.

Comparing `(y-2)^2=4*1/16(x-3)` to the general equation, we get

`h=3, k=2` and `a=1/16`

`implies` `Vertex=(3,2)`

The coordinates of focus of a parabola along x-axis are given by `(h+a,k)`

`implies Focus=(3+1/16,2)=(49/16,2)`

Hence, the coordinates of focus of the given parabola are `(49/16,2).`