When the origin is shifted to the point $(2,3)$ and then the coordinate axes are rotated through an angle…

1. $3 x^2+3 y^2-1=0$
2. $(6+\sqrt{3}) x^2-2 x y+(6-\sqrt{3}) y^2-2=0$
3. $4 x^2+2 y^2-1=0$
4. $(6-\sqrt{3}) x^2+(6+\sqrt{3}) y^2+2 x y=0$

Solution

We first change the origin, by putting $x^{\prime}+2$ and $y^{\prime}+2$ for $x$ and $y$ respectively. The new equation will be $ \begin{gathered} \Rightarrow 3\left(x^{\prime}+2\right)^2+2\left(x^{\prime}+2\right)\left(y^{\prime}+3\right)+3\left(y^{\prime}+3\right)^2- \\ 18\left(x^{\prime}+2\right)-22\left(y^{\prime}+3\right)+50=0 \end{gathered} $ It reduces to
Dropping the suffices, we get the equation $ 3 x^2+2 x y+3 y^2-1=0 $ To turn the axes through an angle of $\frac{\pi}{3}$, in the counter clockwise sense, we must write $\frac{x^{\prime}-\sqrt{3} y^{\prime}}{2}$ for $x$ and $\frac{\sqrt{3} x^{\prime}+y^{\prime}}{2}$ for $y$. The Eq. (i) will then be $ \begin{aligned} & 3\left(\frac{x^{\prime}-\sqrt{3} y^{\prime}}{2}\right)^2+2\left(\frac{x^{\prime}-\sqrt{3} y^{\prime}}{2}\right)\left(\frac{\sqrt{3} x^{\prime}+y^{\prime}}{2}\right)+ \\ & 3\left(\frac{\sqrt{3} x^{\prime}+y^{\prime}}{2}\right)^2-1=0 \end{aligned} $ By dropping the suffices $ \begin{aligned} & 3(x-\sqrt{3} y)^2+2(x-\sqrt{3} y)(\sqrt{3} x+y)+ \\ & 3(\sqrt{3} x+y)^2-4=0 \\ & \Rightarrow 3\left(x^2+3 y^2-2 \sqrt{3} x y\right)+2\left(\sqrt{3} x^2-2 x y-\sqrt{3} y^2\right) \\ &+3\left(3 x^2+y^2+2 \sqrt{3} x y\right)-4=0 \\ & \Rightarrow(12+2 \sqrt{3}) x^2-4 x y+(12-2 \sqrt{3}) y^2-4=0 \\ & \Rightarrow \quad(6+\sqrt{3}) x^2-2 x y+(6-\sqrt{3}) y^2-2=0 \end{aligned} $

Asked in: AP EAMCET 2018 (22 Apr Shift 2)

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