Consider the equation:2sqrt(3x)^2-6xy+sqrt(3x)+3y=0 a) Use the discriminant to determine

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

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