Question

How do you find the vertex, focus and directrix of `x^2-4x+y+3=0`?

Answer 1

Vertex `(2,1)`

Focus `(2,3/4)`

Directrix `y = 5/4`

Explanation:

Complete the square.

`y = -x^2 + 4x -3`

`= -(x -2)^2 + 1`

The vertex is at `(2,1)`.

The parabola is concave down, as the leading coefficient

`a = -1 < 0`

graph{-x^2+4x-3 [-10, 10, -5, 5]}

Therefore, the focus lies `p = 1/(4abs(a))` below the vertex.

`p = frac{1}{4abs(-1)} = 1/4`

The `y`-coordinate of the focus is `1 - 1/4 = 3/4`.

The coordinates of the focus is `(2,3/4)`.

All points on the parabola are equidistant from the focus and the perpendicular to the directrix.

We know that the directrix is a horizontal line, with the equation `y = m`, where `m` is the constant to be determined.

We consider the distance of the vertex to the focus, it is `p = 1/4`. Therefore, the directrix is a horizontal line `1/4` above the vertex `(2,1)`.

`m = 1 + 1/4 = 5/4`

The equation of the directrix is `y = 5/4`.