Chapter A5.3, Problem 22E

To determine

To identify: the graph of the equation.

Answer to Problem 22E

The graph of the equation isparallel horizontal lines

  $4{\left(y^{\prime }\right)}^2=1$

Explanation of Solution

Given:

  $3x^2-2\sqrt{3}xy+y^2=1$

Calculation:

Given the following equation:

  $3x^2-2\sqrt{3}xy+y^2=1$

Comparing with the standard equation for a Quadratic Curve,

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$

We get

  $A=3,B=-2\sqrt{3},C=1$

For the cross-product term to vanish, the coordinate axes must be rotated by an angle alpha,such that

  $\cot 2\alpha =\frac{A-C}{B}$

  $=\frac{3-1}{-2\sqrt{3}}$

  $=\frac{-1}{\sqrt{3}}$

Therefore,

  $2\alpha =\frac{2\pi }{3}$

  $\Rightarrow \alpha =\frac{\pi }{3}$

After rotation of coordinate axes, the original coordinates, $x$ and $y,$ are replaced by newcoordinates, $x^{\prime }$ and $y^{\prime }$ , such that

  $\begin{array}{l}x=x^{\prime }\cos \alpha -y^{\prime }\sin \alpha \\ =\frac{1}{2}{x}^{\prime }-\frac{\sqrt{3}}{2}{y}^{\prime }\\ y=x^{\prime }\sin \alpha +y^{\prime }\cos \alpha \\ =\frac{\sqrt{3}}{2}{x}^{\prime }+\frac{1}{2}{y}^{\prime }\end{array}$

Substituting the above values of $x$ and $y$ in the given Quadratic Equation, $3{\left(\frac{1}{2}{x}^{\prime }-\frac{\sqrt{3}}{2}{y}^{\prime }\right)}^2-2\sqrt{3}\left(\frac{1}{2}{x}^{\prime }-\frac{\sqrt{3}}{2}{y}^{\prime }\right)\left(\frac{\sqrt{3}}{2}{x}^{\prime }+\frac{1}{2}{y}^{\prime }\right)+{\left(\frac{\sqrt{3}}{2}{x}^{\prime }+\frac{1}{2}{y}^{\prime }\right)}^2=1$

  $\Rightarrow 3\left(\frac{1}{4}{\left(x^{\prime }\right)}^2-\frac{\sqrt{3}}{2}{x}^{\prime }{y}^{\prime }+\frac{3}{4}{\left(y^{\prime }\right)}^2\right)$

  $-2\sqrt{3}\left(\frac{\sqrt{3}}{4}{\left(x^{\prime }\right)}^2-\frac{3}{4}{x}^{\prime }{y}^{\prime }+\frac{1}{4}{x}^{\prime }{y}^{\prime }-\frac{\sqrt{3}}{4}{\left(y^{\prime }\right)}^2\right)+\left(\frac{3}{4}{\left(x^{\prime }\right)}^2+\frac{\sqrt{3}}{2}{x}^{\prime }{y}^{\prime }+\frac{1}{4}{\left(y^{\prime }\right)}^2\right)=1$

  $\Rightarrow 4{\left(y^{\prime }\right)}^2=1$

Which are parallel horizontal lines.

Conclusion:

The graph of the equationare parallel horizontal lines

  $4{\left(y^{\prime }\right)}^2=1$