Rui Baldwin

## Answered question

2020-10-25

Consider the equation:
$2\sqrt{3}{x}^{2}-6xy+\sqrt{3}x+3y=0$
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 $0\le \phi \le \pi \text{/}2$)

### Answer & Explanation

Caren

Skilled2020-10-26Added 96 answers

Step 1
We have given an equation,
$2\sqrt{3}{x}^{2}-6xy+\sqrt{3}x+3y=0$
the general Cartesian form of a conic section:
$A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$
The discriminant is ${B}^{2}-4AC$
(a) If ${B}^{2}-4AC<0$, then the equation represents an ellipse.
(b) If ${B}^{2}-4AC=0$, then the equation represents a parabola.
(c) If ${B}^{2}-4AC>0$, then the equation represents a hyperbola.
Step 2
On comparing the given equation with the general form of conic section, we get,
$A=2\sqrt{3},B=-6,C=0$
The discriminant is $-{6}^{2}-4×2\sqrt{3}×0=36$
$36>0$
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:
$x={x}^{\prime }\mathrm{cos}\theta -{y}^{\prime }\mathrm{sin}\theta \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y={x}^{\prime }\mathrm{cos}\theta +{y}^{\prime }\mathrm{sin}\theta$
$\mathrm{cot}2\theta =\frac{C-A}{B}=\frac{0-2\sqrt{3}}{-6}$
$\mathrm{cot}2\theta =\frac{C-A}{B}=\frac{1}{\sqrt{3}}$
$\mathrm{cot}2\theta =\frac{C-A}{B}=\mathrm{cot},\frac{\pi }{3}$
$\theta =\frac{\pi }{6}$
$\mathrm{sin},\frac{\pi }{6}=\frac{1}{2}$
$\mathrm{cos},\frac{\pi }{6}=\frac{\sqrt{3}}{2}$
We shall find the value of x,y by these values.
$x={x}^{\prime }\mathrm{cos}\theta -{y}^{\prime }\mathrm{sin}\theta \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y={x}^{\prime }\mathrm{cos}\theta +{y}^{\prime }\mathrm{sin}\theta$
$x={x}^{\prime }\frac{\sqrt{3}}{2}-{y}^{\prime }\frac{1}{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y={x}^{\prime }\frac{\sqrt{3}}{2}+{y}^{\prime }\frac{1}{2}$
On plugging these values in the equation we get,
$2\sqrt{3}{x}^{2}-6xy+\sqrt{3}x+3y=0$
$2\sqrt{3}{\left({x}^{\prime }\frac{\sqrt{3}}{2}-{y}^{\prime }\frac{1}{2}\right)}^{2}-6\left({x}^{\prime }\frac{\sqrt{3}}{2}-{y}^{\prime }\frac{1}{2}\right)\left({x}^{\prime }\frac{\sqrt{3}}{2}+{y}^{\prime }\frac{1}{2}\right)+\sqrt{3}\left({x}^{\prime }\frac{\sqrt{3}}{2}-{y}^{\prime }\frac{1}{2}\right)+3\left({x}^{\prime }\frac{\sqrt{3}}{2}+{y}^{\prime }\frac{1}{2}\right)=0$

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