Chapter 10, Problem 40RE

To determine

Tofind:the conic without completing the square and rotational axes.

Answer to Problem 40RE

The equation is hyperbolic equation.

Explanation of Solution

Given:

  ${\text{x}}^{\text{2}}{\text{-8y}}^{\text{2}}\text{-x-2y=0}$ .

Concept used:

The general equation of conic:

  ${\text{Ax}}^{\text{2}}{\text{+Bxy+Cy}}^{\text{2}}\text{+Dx+Ey+F=0}$

Defines:

A parabola if ${\text{B}}^{\text{2}}\text{-4AC=0}$ .

An ellipse if ${\text{B}}^{\text{2}}\text{-4AC<0}$ $\text{and}$ $\text{A}\ne \text{B}$ .

A circle if ${\text{B}}^{\text{2}}\text{-4AC<0}$ $\text{and}$ $\text{A=B}$ .

A hyperbola if ${\text{B}}^{\text{2}}\text{-4AC>0}$ .

The general equation of conic if $\text{B=0}$ $\text{and}$ $\text{A and B}$ are not both zero.

  ${\text{Ax}}^{\text{2}}{\text{+Cy}}^{\text{2}}\text{+Dx+Ey+F=0}$

Defines:

A parabola if $\text{AC=0}$ .

Here either $\text{A=0 or B=0}$ but not both.

An ellipse if $\text{AC>0}$ $\text{and}$ $\text{A}\ne \text{B}$ .

That is $\text{A and C}$ have same sign, but $\text{A}\ne \text{C}$ .

A circle if $\text{AC>0}$ and $\text{A=C}$

That is $\text{A=C}$ .

A hyperbola if $\text{AC<0}$ .

That is $\text{A}$ $\text{and C}$ have opposite signs.

Calculation:

Here $\text{A=1}$ , $\text{B=0}$ and $\text{C=-8}$ .

According to the formula:

A hyperbola if $\text{AC<0}$ .

That is $\text{A}$ $\text{and C}$ have opposite signs.

It satisfies the equation of hyperbola.

Hence the equation is hyperbolic equation.