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(B) The rotation of axes by an angle $	heta = 30^{\circ}$ is given by the transformation: $x = X \cos 	heta - Y \sin 	heta = X \frac{\sqrt{3}}{2} -

Vedclass · Accepted Answer

$g/f=\frac{3+2\sqrt{3}}{3\sqrt{3}-2}$

Vedclass · Answer

(B) The rotation of axes by an angle $	heta = 30^{\circ}$ is given by the transformation: $x = X \cos 	heta - Y \sin 	heta = X \frac{\sqrt{3}}{2} - Y \frac{1}{2}$ $y = X \sin 	heta + Y \cos 	heta = X \frac{1}{2} + Y \frac{\sqrt{3}}{2}$ Substituting these into the equation $4x^2+12xy+9y^2+6x+9y+2=0$: Note that $4x^2+12xy+9y^2 = (2x+3y)^2$. Substituting $x$ and $y$: $2x+3y = 2(\frac{\sqrt{3}}{2}X - \frac{1}{2}Y) + 3(\frac{1}{2}X + \frac{\sqrt{3}}{2}Y) = X(\sqrt{3} + \frac{3}{2}) + Y(\frac{3\sqrt{3}-2}{2})$. Squaring this gives the $X^2, XY, Y^2$ terms. The linear part is $6x+9y = 6(\frac{\sqrt{3}}{2}X - \frac{1}{2}Y) + 9(\frac{1}{2}X + \frac{\sqrt{3}}{2}Y) = X(3\sqrt{3} + \frac{9}{2}) + Y(\frac{9\sqrt{3}-6}{2})$. Thus,$2g = \frac{6\sqrt{3}+9}{2}$ and $2f = \frac{9\sqrt{3}-6}{2}$. Calculating the ratio: $\frac{g}{f} = \frac{6\sqrt{3}+9}{9\sqrt{3}-6} = \frac{3(2\sqrt{3}+3)}{3(3\sqrt{3}-2)} = \frac{3+2\sqrt{3}}{3\sqrt{3}-2}$. Therefore,option $B$ is correct.

By rotating the axes through an angle of $30^{\circ}$ in the anti-clockwise direction about the origin,the equation $4x^2+12xy+9y^2+6x+9y+2=0$ becomes $ax^2+2hxy+by^2+2gx+2fy+c=0$. Then which of the following is true?

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