Chapter 8.1, Problem 52E

Solution

To prove: The graph of the equation is a parabola, and find its vertex, focus and directrix.

The vertex is $\left(\frac{13}{4},1\right)$ .

The focus is $\left(\frac{9}{4},1\right)$ .

The directrix is $\frac{17}{4}$ .

Given information: The graph of the equation is a parabola, and find its vertex, focus and directrix.

  $y^2-2y+4x-12=0$

Concept used: Use the equation of parabola in the standard form to find its vertex, focus and directrix.

Explanation:

Since the equation is quadratic in variable $y$ , complete the square with respect to $y$ to obtain the standard form.

  $\begin{array}{l}{y}^2-2y+4x+12=0\\ y^2-2y=-4x+12\left[\text{Isolate}\text{the}y\mathrm{term}\right]\\ y^2-2y+1=-4x+12+1\left[\text{Complete}\text{the}\text{square}\right]\\ {\left(y-1\right)}^2=-4x+13\\ {\left(y-1\right)}^2=-4\left(x-\frac{13}{4}\right)\end{array}$

This equation is in standard form:

  ${\left(y-k\right)}^2=4p\left(x-h\right)$

where $h=\frac{13}{4},k=1,p=-1$ .

It follows that:

The graph of the equation is a parabola.

The vertex $\left(h,k\right)$ is $\left(\frac{13}{4},1\right)$ .

The focus $\left(h+p,k\right)$ will be:

  $\begin{array}{l}\left(\frac{13}{4}-1,1\right)\\ =\left(\frac{9}{4},1\right)\end{array}$

The directrix will be:

  $\begin{array}{l}x=h-p\\ =\frac{13}{4}-\left(-1\right)\\ =\frac{13}{4}+1\\ =\frac{17}{4}\end{array}$