91Ó°ÊÓ

Given to identify the equation and

• if it is a parabola, its vertex, focus, and directrix;
• if an ellipse, its center, vertices, and foci;
• if a hyperbola, its center, vertices, foci, and asymptotes.

are to be determined.

Given equation is$8y={\left(x-1\right)}^2-4$

Rewriting the given equation:

$$\begin{gathered} 8y={\left(x-1\right)}^2-4 \\ {\left(x-1\right)}^2=4\left(2\right)\left(y+\frac{1}{2}\right) \end{gathered}$$

The equation is of the form ${\left(x-h\right)}^2=4a\left(y-k\right)$

It is a parabola with vertex at $\left(h,k\right)$

The focus is $\left(h,k+a\right)$

The directrix is $y=k-a$

Hence the conic is a parabola.

In the given equation, $h=1,k=-\frac{1}{2},a=2$

Plugging the values to determine vertex, focus and asymptote:

The vertex of the parabola is $\left(h,k\right)=\left(1,-\frac{1}{2}\right)$

The focus of the parabola is given by:

$\begin{array}{rcl}\left(h,k+a\right)& =& \left(1,-\frac{1}{2}+2\right)\\ & =& \left(1,\frac{3}{2}\right)\end{array}$

The directrix of the parabola is given by:

$$\begin{gathered} y=k-a \\ y=-\frac{1}{2}-2 \\ y=-\frac{5}{2} \end{gathered}$$

The equation is a parabola with vertex at $\left(1,-\frac{1}{2}\right)$and the focus is$\left(1,\frac{3}{2}\right)$and the directrix is$y=-\frac{5}{2}$