Question

Consider the equation: displaystyle{2}sqrt{{3}}{x}^{2}-{6}{x}{y}+sqrt{{3}}{x}+{3}{y}={0} a) Use the discriminant to determine whether the graph of the

Conic sections
Consider the equation:
$$\displaystyle{2}\sqrt{{3}}{x}^{2}-{6}{x}{y}+\sqrt{{3}}{x}+{3}{y}={0}$$
a) Use the discriminant to determine whether the graph of the equation is a parabola, an ellipse, or a hyperbola?
b) Use a rotation of axes to eliminate the xy-term. (Write an equation in XY-coordinates. Use a rotation angle that satisfies $$\displaystyle{0}\le\varphi\le\pi\text{/}{2}$$)

2020-10-26

Step 1
We have given an equation,
$$\displaystyle{2}\sqrt{{3}}{x}^{2}-{6}{x}{y}+\sqrt{{3}}{x}+{3}{y}={0}$$
the general Cartesian form of a conic section:
$$\displaystyle{A}{x}^{2}+{B}{x}{y}+{C}{y}^{2}+{D}{x}+{E}{y}+{F}={0}$$
The discriminant is $$\displaystyle{B}^{2}-{4}{A}{C}$$
(a) If $$\displaystyle{B}^{2}-{4}{A}{C}<{0}$$, then the equation represents an ellipse.
(b) If $$\displaystyle{B}^{2}-{4}{A}{C}={0}$$, then the equation represents a parabola.
(c) If $$\displaystyle{B}^{2}-{4}{A}{C}>{0}$$, then the equation represents a hyperbola.
Step 2
On comparing the given equation with the general form of conic section, we get,
$$\displaystyle{A}={2}\sqrt{{3}},{B}=-{6},{C}={0}$$
The discriminant is $$-\displaystyle{6}^{2}-{4}\times{2}\sqrt{{3}}\times{0}={36}$$
$$\displaystyle{36}>{0}$$
So the equation is hyperbola.
Step 3
b) In such a case, the relation between coordinate (x, y) and new coordinates (x', y') is given by:
$$\displaystyle{x}={x}' \cos{\theta}-{y}' \sin{\theta}{\quad\text{and}\quad}{y}={x}' \cos{\theta}+{y}' \sin{\theta}$$
$$\displaystyle \cot{{2}}\theta=\frac{{{C}-{A}}}{{B}}=\frac{{{0}-{2}\sqrt{{3}}}}{ -{{6}}}$$
$$\displaystyle \cot{{2}}\theta=\frac{{{C}-{A}}}{{B}}=\frac{1}{\sqrt{{3}}}$$
$$\displaystyle \cot{{2}}\theta=\frac{{{C}-{A}}}{{B}}={\cot},\frac{\pi}{{3}}$$
$$\displaystyle\theta=\frac{\pi}{{6}}$$
$$\displaystyle{\sin},\frac{\pi}{{6}}=\frac{1}{{2}}$$
$$\displaystyle{\cos},\frac{\pi}{{6}}=\frac{\sqrt{{3}}}{{2}}$$
We shall find the value of x,y by these values.
$$\displaystyle{x}={x}' \cos{\theta}-{y}' \sin{\theta}{\quad\text{and}\quad}{y}={x}' \cos{\theta}+{y}' \sin{\theta}$$
$$\displaystyle{x}={x}'\frac{\sqrt{{3}}}{{2}}-{y}'\frac{1}{{2}}{\quad\text{and}\quad}{y}={x}'\frac{\sqrt{{3}}}{{2}}+{y}'\frac{1}{{2}}$$
On plugging these values in the equation we get,
$$\displaystyle{2}\sqrt{{3}}{x}^{2}-{6}{x}{y}+\sqrt{{3}}{x}+{3}{y}={0}$$
$$\displaystyle{2}\sqrt{{3}}{\left({x}'\frac{\sqrt{{3}}}{{2}}-{y}'\frac{1}{{2}}\right)}^{2}-{6}{\left({x}'\frac{\sqrt{{3}}}{{2}}-{y}'\frac{1}{{2}}\right)}{\left({x}'\frac{\sqrt{{3}}}{{2}}+{y}'\frac{1}{{2}}\right)}+\sqrt{{3}}{\left({x}'\frac{\sqrt{{3}}}{{2}}-{y}'\frac{1}{{2}}\right)}+{3}{\left({x}'\frac{\sqrt{{3}}}{{2}}+{y}'\frac{1}{{2}}\right)}={0}$$
$$\displaystyle\frac{\sqrt{{3}}}{{2}}{\left(\sqrt{{3}}{x}'-{y}'\right)}^{2}-\frac{3}{{2}}{\left(\sqrt{{3}}{x}'-{y}'\right)}{\left(\sqrt{{3}}{x}'+{y}'\right)}+\frac{\sqrt{{3}}}{{2}}{\left(\sqrt{{3}}{x}'-{y}'\right)}+\frac{3}{{2}}{\left(\sqrt{{3}}{x}'+{y}'\right)}={0}$$
$$\displaystyle\frac{\sqrt{{3}}}{{2}}{\left({3}{x}'^{2}+{y}'^{2}-{2}\sqrt{{3}}{x}'{y}'\right)}-\frac{3}{{2}}{\left({3}{x}'^{2}-{y}'^{2}\right)}+\frac{\sqrt{{3}}}{{2}}{\left(\sqrt{{3}}{x}'-{y}'\right)}+\frac{3}{{2}}{\left(\sqrt{{3}}{x}'+{y}'\right)}={0}$$
$$\displaystyle\frac{\sqrt{{3}}}{{2}}{\left({3}{x}'^{2}+{y}'^{2}-{2}\sqrt{{3}}{x}'{y}'\right)}-\frac{3}{{2}}{\left({3}{x}'^{2}-{y}'^{2}\right)}+\frac{\sqrt{{3}}}{{2}}{\left(\sqrt{{3}}{x}'-{y}'\right)}+\frac{3}{{2}}{\left(\sqrt{{3}}{x}'+{y}'\right)}={0}$$
$$\displaystyle\frac{1}{{2}}{\left[\sqrt{{3}}{\left({3}{x}'^{2}+{y}'^{2}-{2}\sqrt{{3}}{x}'{y}'\right)}-{3}{\left({3}{x}'^{2}-{y}'^{2}\right)}+\sqrt{{3}}{\left(\sqrt{{x}}'-{y}'\right)}+{3}{\left(\sqrt{{3}}{x}'+{y}'\right)}\right]}={0}$$
$$\displaystyle{\left[{3}\sqrt{{3}}{x}'^{2}+\sqrt{{3}}{y}'^{2}-{6}{x}'{y}'-{9}{x}'^{2}+{3}{y}'^{2}+{3}{x}'-\sqrt{{3}}{y}'+{3}\sqrt{{3}}{x}'+{3}{y}'\right]}={0}$$
$$\displaystyle{\left({3}\sqrt{{3}}-{9}\right)}{x}'^{2}+{\left(\sqrt{{3}}+{3}\right)}{y}'^{2}-{6}{x}'{y}'+{\left({3}+{3}\sqrt{{3}}\right)}{x}'-{\left(\sqrt{{3}}-{3}\right)}{y}'={0}$$