When the origin is shifted to \((2,3)\) the transformed equation \(x^2+3 x y-2 y^2+17 x-7 y-11=0\), then the…

When the origin is shifted to \((2,3)\) the transformed equation \(x^2+3 x y-2 y^2+17 x-7 y-11=0\), then the original equation of curve is
  1. \(x^2-2 y^2-3 x y+4 x-y+20=0\)
  2. \(x^2-2 y^2+3 x y+4 x-y-20=0\)
  3. \(x^2-2 y^2-3 x y-4 x-y+20=0\)
  4. \(x^2-2 y^2-3 x y+4 x-y-20=0\)

Solution

It is given that the origin is shifted to point \((2,3)\) and due to that the transformed equation of the curve is, \(x^2+3 x y-2 y^2+17 x-7 y-11=0\), to get the original equation of curve, replace \((x, y)\) by \((x-2, y-3)\), so by doing this. we get, \((x-2)^2+3(x-2)(y-3)-2(y-3)^2\) \(\begin{gathered} +17(x-2)-7(y-3)-11=0 \\ \Rightarrow x^2-4 x+4+3(x y-3 x-2 y+6)-2\left(y^2-6 y+9\right) \\ +17 x-34-7 y+21-11=0 \\ \Rightarrow x^2+3 x y-2 y^2+4 x-y-20=0 \end{gathered}\) Hence, option (b) is correct.

Asked in: AP EAMCET 2020 (18 Sep Shift 2)

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