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

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Aicharya · Accepted Answer

Given, equation is $3 x^2+3 y^2+2 x y=2 \ldots$ (i) When coordinate axis are rotated through an angle $	heta$, then $ x=X \cos 	heta-Y \sin 	heta 	ext { and } y=X \sin 	heta+Y \cos 	heta $ where, $(X, Y)$ is new coordinate after transformation. Given, $	heta=45^{\circ}$, then $x=\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}}$ and $y=\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}$ Putting value of $x$ and $y$ in Eq. (i), we get $ \begin{aligned} & 3\left(\frac{X}{\sqrt{2}}-\frac{Y}{\sqrt{2}}ight)^2+3\left(\frac{X}{\sqrt{2}}+\frac{Y}{\sqrt{2}}ight)^2 \ & +2\left(\frac{X}{\sqrt{2}}-\frac{Y}{\sqrt{2}}ight)\left(\frac{X}{\sqrt{2}}+\frac{Y}{\sqrt{2}}ight)=2 \ & \Rightarrow \frac{3}{2}(X-Y)^2+\frac{3}{2}(X+Y)^2+\left(X^2-Y^2ight)=2 \ & \Rightarrow \quad \frac{3}{2}\left[2 X^2+2 Y^2ight]+X^2-Y^2=2 \ & \Rightarrow \quad 3 X^2+3 Y^2+X^2-Y^2=2 \ & \Rightarrow \quad 4 X^2+2 Y^2=2 \ & 	ext { or } \ & 2 X^2+Y^2=1 \ & \end{aligned} $ $	herefore$ Transformed equation is $2 x^2+y^2=1$

The transformed equation $3 x^2+3 y^2+2 x y=2$, when the coordinate axes are rotated through an angle…

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