Chapter 10, Problem 2CT

To determine

The identification of the equation $8y={(x-1)}^2-4$ . If it is a parabola, give a vertex, focus, and directrix; if it is an ellipse, give its center, vertices, and foci; if it is a hyperbola, give its center, vertices, foci, and asymptotes.

Answer to Problem 2CT

Solution:

The equation $8y={(x-1)}^2-4$ represents equation of parabola, its vertex is $\left(1,-\frac{1}{2}\right)$, focus is $\left(1,\frac{3}{2}\right)$, and directrix is $y=-\frac{5}{2}$ .

Explanation of Solution

Given information:

The equation $8y={(x-1)}^2-4$

Consider $8y={(x-1)}^2-4$ ,

Rewrite the given equation by adding $4$ to both sides,

  $\Rightarrow 8y+4={(x-1)}^2$

Factor out $8$ from the left side,

  $\Rightarrow 8\left(y+\frac{1}{2}\right)={(x-1)}^2$

  $\Rightarrow {(x-1)}^2=8\left(y+\frac{1}{2}\right)$

The equation ${(x-1)}^2=8\left(y+\frac{1}{2}\right)$ is of the form ${\left(x-h\right)}^2=4a\left(y-k\right)$ .

Hence, the equation $8y={(x-1)}^2-4$ represents a parabola.

The axis of symmetry is parallel to $y$ -axis.

If the equation of the parabola is of the form ${\left(x-h\right)}^2=4a\left(y-k\right)$, then its vertex is $\left(h,k\right)$, focus $\left(h,k+a\right)$, and directrix is $y=k-a$ .

Compare the equation ${(x-1)}^2=8\left(y+\frac{1}{2}\right)$ with ${\left(x-h\right)}^2=4a\left(y-k\right)$, get $h=1,k=-\frac{1}{2},$

  $a=2$ .

Vertex of the given equation $\left(h,k\right)=\left(1,-\frac{1}{2}\right)$, focus is $\left(h,k+a\right)=\left(1,-\frac{1}{2}+2\right)=\left(1,\frac{3}{2}\right)$, and directrix is $y=k-a\Rightarrow y=-\frac{1}{2}-2=-\frac{5}{2}$ .

Thus, $x^2-4x=2y$ is represents the equation of parabola, its vertex is $\left(1,-\frac{1}{2}\right)$, focus is $\left(1,\frac{3}{2}\right)$, and directrix is $y=-\frac{5}{2}$ .