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

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(B) Since the axes are rotated through an angle $	heta = 45^{\circ}$,we replace $(x, y)$ by $(x \cos 45^{\circ} - y \sin 45^{\circ}, x \sin 45^{\circ

Vedclass · Accepted Answer

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

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(B) Since the axes are rotated through an angle $	heta = 45^{\circ}$,we replace $(x, y)$ by $(x \cos 45^{\circ} - y \sin 45^{\circ}, x \sin 45^{\circ} + y \cos 45^{\circ})$,which is $(\frac{x-y}{\sqrt{2}}, \frac{x+y}{\sqrt{2}})$. Substituting these into the equation $3x^2 + 3y^2 + 2xy = 2$:$3(\frac{x-y}{\sqrt{2}})^2 + 3(\frac{x+y}{\sqrt{2}})^2 + 2(\frac{x-y}{\sqrt{2}})(\frac{x+y}{\sqrt{2}}) = 2$$\frac{3}{2}(x^2 + y^2 - 2xy) + \frac{3}{2}(x^2 + y^2 + 2xy) + \frac{2}{2}(x^2 - y^2) = 2$$\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 $2x^2 + y^2 = 1$.

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

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