Chapter 10, Problem 46RE

To determine

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

Answer to Problem 46RE

The equation is hyperbolic equation.

Explanation of Solution

Given:

  ${\text{4x}}^{\text{2}}{\text{+12xy-10y}}^{\text{2}}\text{+x+y-10=0}$ .

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:

According to the given:

  ${\text{4x}}^{\text{2}}{\text{+12xy-10y}}^{\text{2}}\text{+x+y-10=0}$

Here $\text{A=4}$ , $\text{B=12}$ and $\text{C=-10}$ .

Thus, $\text{B}\ne \text{0}$

According to the formula:

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

  ${\left(\text{12}\right)}^{\text{2}}\text{-4}\left(\text{4}\right)\left(\text{-10}\right)\text{=304>0}$ .

By putting the value of $\text{A,B and C}$ .

It satisfies the equation of hyperbola

Hence the equation is hyperbolic equation.