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

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Answer 1 · 2016-04-21T05:23:59

Vertex $(2,1)$

Focus $(2,3/4)$

Directrix $y = 5/4$

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.

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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 $y = 5/4$ .

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