Question

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

• A. $x^2 + 2y^2 = 1$
• B. $2x^2 + y^2 = 1$
• C. $x^2 + y^2 = 1$
• D. $x^2 + 3y^2 = 1$

Answer 1

Answer: (B)

When axes are rotated through $45^{\circ}$,
$x=x \cos \theta-y \sin \theta=\frac{x-y}{\sqrt{2}}$
$y=x \sin \theta-y \cos \theta=\frac{x+y}{\sqrt{2}}$
$3 x^{2}+3 y^{2}+2 x y=2$
$\Rightarrow 3\left(\frac{x-y}{\sqrt{2}}\right)^{2}+3\left(\frac{x +y}{\sqrt{2}}\right)^{2}+2\left(\frac{x^{2}-y^{2}}{2}\right)=2$
$3\left(x^{2}+y^{2}-2 x y\right)+3\left(x^{2}+y^{2}+2 x y\right)+2 x^{2}-2 y^{2}=4$
$4\left(2 x^{2}+y^{2}\right)=4$
$\Rightarrow 2 x^{2}+y^{2}=1$