Q86 / 160 · TS EAMCET 2008 · Mathematics
The transformed equation of $3 x^2+3 y^2+2 x y=2$, when the coordinate axes are rotated through an angle of $45^{\circ}$, is
- A$x^2+2 y^2=1$
- Correct answerB$2 x^2+y^2=1$
- C$x^2+y^2=1$
- D$x^2+3 y^2=1$
Solution
Since, the axes are rotated through an angle $45^{\circ}$, then we replace $(x, y)$ by $\left(x \cos 45^{\circ}-y \sin 45^{\circ}, x \sin 45^{\circ}+y \cos 45^{\circ}\right)$ ie, $\quad\left(\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}}, \frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}\right)$ in the given equation $3 x^2+3 y^2+2 x y=2$
$$
\begin{array}{rrr}
\therefore \quad 3\left(\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}}\right)^2+3\left(\frac{x+y}{\sqrt{2}}\right)^2 \\
& +2\left(\frac{x-y}{\sqrt{2}}\right)\left(\frac{x+y}{\sqrt{2}}\right)=2 \\
\Rightarrow & \frac{3}{2}\left(x^2+y^2+2 x y\right)+\frac{3}{2}\left(x^2+y^2-2 x y\right) \\
& & +\frac{2}{2}\left(x^2-y^2\right)=2 \\
\Rightarrow & 4 x^2+2 y^2=2 \\
\Rightarrow & 2 x^2+y^2=1
\end{array}
$$
Hence, the correct answer is option: 'B'.