The transformed equation 3x^2 + 3y^2 + 2xy = | vedclass

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(B) Given equation is $3x^2 + 3y^2 + 2xy = 2 \dots (i)$.When coordinate axes are rotated through an angle $	heta = 45^{\circ}$,the transformation equ

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$2x^2 + y^2 = 1$

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(B) Given equation is $3x^2 + 3y^2 + 2xy = 2 \dots (i)$.When coordinate axes are rotated through an angle $	heta = 45^{\circ}$,the transformation equations are:$x = X \cos 45^{\circ} - Y \sin 45^{\circ} = \frac{X - Y}{\sqrt{2}}$$y = X \sin 45^{\circ} + Y \cos 45^{\circ} = \frac{X + Y}{\sqrt{2}}$Substituting these into equation $(i)$:$3\left(\frac{X - Y}{\sqrt{2}}ight)^2 + 3\left(\frac{X + Y}{\sqrt{2}}ight)^2 + 2\left(\frac{X - Y}{\sqrt{2}}ight)\left(\frac{X + Y}{\sqrt{2}}ight) = 2$$\frac{3}{2}(X^2 - 2XY + Y^2) + \frac{3}{2}(X^2 + 2XY + Y^2) + (X^2 - Y^2) = 2$$\frac{3}{2}(2X^2 + 2Y^2) + X^2 - Y^2 = 2$$3X^2 + 3Y^2 + X^2 - Y^2 = 2$$4X^2 + 2Y^2 = 2$Dividing by $2$,we get $2X^2 + Y^2 = 1$.Thus,the transformed equation is $2x^2 + y^2 = 1$.

The transformed equation $3x^2 + 3y^2 + 2xy = 2$,when the coordinate axes are rotated through an angle $45^{\circ}$,is

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