Chapter 10, Problem 42RE

To determine

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

Answer to Problem 42RE

The equation is parabolic equation.

Explanation of Solution

Given:

  ${\text{4x}}^{\text{2}}{\text{+4xy+y}}^{\text{2}}\text{-8}\sqrt{\text{5}}\text{x+16}\sqrt{\text{5}}\text{y=0}$ .

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:

Given equation is ${\text{4x}}^{\text{2}}{\text{+4xy+y}}^{\text{2}}\text{-8}\sqrt{\text{5}}\text{x+16}\sqrt{\text{5}}\text{y=0}$ .

Here $\text{A=4}$ , $\text{B=4}$ and $\text{C=1}$ .

  ${\text{B}}^{\text{2}}\text{-4AC=}{\left(4\right)}^2-4\left(4\right)\left(1\right)=0$ .

According to the formula:

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

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

It satisfies the equation of parabola

Hence the equation is parabolic equation.