Chapter 10.7, Problem 33E

To determine

Tofind:the graph of the equation and the graph of equation is degenerated case.

Answer to Problem 33E

The new equation so form is $y=-2x-3$ .

Explanation of Solution

Given:

  ${\left(x-2\right)}^2-{\left(x+3\right)}^2=5\left(y+2\right)$ .

Concept used:

The general equation for conic section is:

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0.$

Where $A,B,C,D,E\text{ and }F$ are constant.

As by changing the values of some of the constants, the shape of the corresponding conic also changes.

If $B^2-4AC$ , is less than zero, if a conic exists, it will be either a circle or an ellipse.

If $B^2-4AC$ ,equals to zero, if a conic exists, it will be parabola.

If $B^2-4AC$ id greater than zero, if a conic exists, it will be hyperbola.

Calculation:

Consider the equation of the conic:

  ${\left(x-2\right)}^2-{\left(x+3\right)}^2=5\left(y+2\right)$ .

Rearrange the equation in the general form of conic:

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0.$

  $B=0$ if conic parallel to co-ordinate axis else $B\ne 0$ .

  $\begin{array}{l}\left(x^2-\mathrm{2.2.}x+4\right)-\left(x^2+\mathrm{2.3.}x+9\right)=5\left(y+2\right).\\ \Rightarrow x^2-4x+4-x^2-6x-9=5\left(y+2\right).\\ \Rightarrow -10x-5=5\left(y+2\right).\\ -2x-y-3=0.\end{array}$

Comparing the equation with general form of conic above:

  $\begin{array}{l}A=0,B=0,C=0,\\ D=-2,E=-1,F=-3.\end{array}$

Now,

  $B^2-4AC=0-\mathrm{4.0.0}=0.$

As in the given equation ${\left(x-2\right)}^2-{\left(x+3\right)}^2=5\left(y+2\right)$ value of $B^2-4AC=0$ so, at first glance the given equation may appear to be that of a parabola but a closer inspection reveals that this is a degenerate case.

So, by solving the equation ${\left(x-2\right)}^2-{\left(x+3\right)}^2=5\left(y+2\right)$ to get:

  $\begin{array}{l}-2x-y-3=0.\\ y=-2x-3.\end{array}$

Now find the certain points to graph the straight line $y=-2x-3$ :

The graph of equation $y=-2x-3$ is:

  

Hence, the new equation so form is $y=-2x-3$ .