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

Apr 21, 2016

Vertex $\left(2 , 1\right)$

Focus $\left(2 , \frac{3}{4}\right)$

Directrix $y = \frac{5}{4}$

#### Explanation:

Complete the square.

$y = - {x}^{2} + 4 x - 3$

$= - {\left(x - 2\right)}^{2} + 1$

The vertex is at $\left(2 , 1\right)$.

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 $p = \frac{1}{4 \left\mid a \right\mid}$ below the vertex.

$p = \frac{1}{4 \left\mid - 1 \right\mid} = \frac{1}{4}$

The $y$-coordinate of the focus is $1 - \frac{1}{4} = \frac{3}{4}$.

The coordinates of the focus is $\left(2 , \frac{3}{4}\right)$.

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 $p = \frac{1}{4}$. Therefore, the directrix is a horizontal line $\frac{1}{4}$ above the vertex $\left(2 , 1\right)$.

$m = 1 + \frac{1}{4} = \frac{5}{4}$

The equation of the directrix is $y = \frac{5}{4}$.