Chapter 10.7, Problem 35E

To determine

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

Answer to Problem 35E

The new equation so form is $y=\pm 3x$ .

Explanation of Solution

Given:

  $y^2-9x^2=0.$

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:

  $y^2-9x^2=0.$

Rearrange the equation in the general form of conic:

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0.$

Rearranging the above equation in this form:

  $-9x^2+y^2=0.$

Comparing the equation with general form of conic above:

  $\begin{array}{l}A=-9,B=0,C=1,\\ D=0,E=0,F=0.\end{array}$

Now,

  $B^2-4AC=0^2-4.\left(-9\right).1=36.$

While $B^2-4AC>0$ ,from the table of comparison of conic, $y^2-9x^2=0$ represents a hyperbola.

Write the equation $-9x^2+y^2=0.$ in the standard form,

  $y^2-9x^2=\mathrm{0........}\left(i\right)$

The standard equation of an ellipse is:

  $\frac{{\left(x-k\right)}^2}{b^2}-\frac{{\left(x-h\right)}^2}{a^2}=\mathrm{1........}\left(ii\right)$ .

Where $\left(h,k\right):\text{center, }a:\text{conjugate-axis,}b:\text{ transverse axis}$ .

Compare the equation $\left(i\right)\text{ and }\left(ii\right)$ :

  $h=0,k=0,a=0,b=0.$

Therefore, the equation $1$ describe a degenerate hyperbola or a pair of lines.

Use equation $1$ to get:

  $y=\pm 3x\mathrm{.......}\left(E\right)$ .

It represents the pair of lines.

The graph of the equation $y^2-9x^2=0.$ is:

  

Hence, the new equation so form is $y=\pm 3x$ .