Chapter A5.3, Problem 20E

To determine

To identify: the graph of the equation.

Answer to Problem 20E

The graph of the equation isan $\text{ellipse}$

  ${\left(x^{\prime }\right)}^2+5{\left(y^{\prime }\right)}^2=2$

Explanation of Solution

Given:

  $x^2-\sqrt{3}xy+2y^2=1$

Calculation:

Given the following equation:

  $x^2-\sqrt{3}xy+2y^2=1$

Comparing with the standard equation for a Quadratic Curve,

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$

Thus,

  $A=1,B=-\sqrt{3},C=2$

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

such that

  $\cot 2\alpha =\frac{A-C}{B}$

  $=\frac{1-2}{-\sqrt{3}}$

  $=\frac{1}{\sqrt{3}}$

Therefore,

  $2\alpha =\frac{\pi }{3}$

  $\Rightarrow \alpha =\frac{\pi }{6}$

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

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

Substituting the above values of $x$ and $y$ in the given Quadratic Equation,

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

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

  $\Rightarrow \frac{1}{2}{\left(x^{\prime }\right)}^2+\frac{5}{2}{\left(y^{\prime }\right)}^2=1$

  $\Rightarrow {\left(x^{\prime }\right)}^2+5{\left(y^{\prime }\right)}^2=2$

Which is an $\text{ellipse}$

Conclusion:

The graph of the equation isan $\text{ellipse}$

  ${\left(x^{\prime }\right)}^2+5{\left(y^{\prime }\right)}^2=2$