Chapter 8, Problem 21RE

Solution

To find: The type of conic, write it in standard form by completing the square and sketch the graph.

The conic is a parabola, the standard form is ${\left(x-3\right)}^2=y+12$ , and the graph is shown in figure (1).

Given information:

The second degree equation is

  $x^2-6x-y-3=0$ .

Formula used:

The second degree equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ graphs as

Hyperbola if $$

  $B^2-4AC>0$ .

Parabola if $B^2-4AC=0$ .

Ellipse if $B^2-4AC<0$ .

Calculation:

Comparing the given equation $x^2-6x-y-3=0$ with $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ gives

  $A=1$

  $B=0$

  $C=0$

Then

  $\begin{array}{c}{B}^2-4AC=0-4\left(1\right)\left(0\right)\\ B^2-4AC=0\end{array}$

Thus, the given conic is a parabola.

Again considering the equation

  $x^2-6x-y-3=0$

Adding $3$ to both sides of this equation

  $x^2-6x-y=3$

Adding $9$ to both sides of this equation

  $\begin{array}{c}{x}^2-6x+9-y=3+9\\ {\left(x-3\right)}^2-y=12\\ {\left(x-3\right)}^2=y+12\end{array}$

This is the standard form of parabola that opens upward or downward.

This equation can rewritten as

  ${\left(x-3\right)}^2=y-\left(-12\right)$   .....  .....….(1)

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

  $\begin{array}{c}4p=1\\ p=\frac{1}{4}\end{array}$

Since $p=\frac{1}{4}>0$ thus this parabola opens upward.

And

  $\left(h,k\right)=\left(3,-12\right)$

Thus, the given parabola has vertex $\left(3,-12\right)$

Substituting $4$ for $y$ in equation (1) gives

  $\begin{array}{c}{\left(x-3\right)}^2=4-\left(-12\right)\\ {\left(x-3\right)}^2=16\\ x-3=\pm 4\\ x=-1,7\end{array}$

Thus $\left(-1,4\right)$ and $\left(7,4\right)$ are two points on the given parabola.

The graph of the equation ${\left(x-3\right)}^2=y+12$ is

  

Therefore, the correct graph of the given equation ${\left(x-3\right)}^2=y+12$ is shown above.