The transformed equation of $2 x^2+3 y^2-z^2-8 x+18 y$ $+2 z+9=0$ when the axes are translated to the point…
The transformed equation of $2 x^2+3 y^2-z^2-8 x+18 y$ $+2 z+9=0$ when the axes are translated to the point $(2,-3,1)$ is
- $2 x^2+3 y^2-z^2=25$
- $2 x^2+3 y^2+z^2=25$
- $2 x^2-3 y^2-z^2=25$
- $2 x^2+3 y^2-z^2=50$
Solution
Since the axes are translated by $(2,-3,1)$
$\begin{aligned} & \text { Let } x-2=X \Rightarrow x=X+2 \\ & y+3=Y \Rightarrow y=Y-3 \\ & z-1=Z \Rightarrow z=Z+1\end{aligned}$
$\because 2 x^2+3 y^2-z^2-8 x+18 y+2 z+9=0$
$\begin{array}{ll}\Rightarrow \quad & 2(\mathrm{X}+2)^2+3(\mathrm{Y}-3)^2-(\mathrm{Z}+1)^2-8(\mathrm{X}+2)+18 \\ & (\mathrm{Y}-3)+2(\mathrm{Z}+1)+9=0 \\ \Rightarrow \quad & 2\left(\mathrm{X}^2+4+4 \mathrm{X}\right)+3\left(\mathrm{Y}^2+9-6 \mathrm{Y}\right)-\left(\mathrm{Z}^2+1+2 \mathrm{Z}\right) \\ & -8 \mathrm{X}-16+18 \mathrm{Y}-54+2 \mathrm{Z}+2+9=0 \\ \Rightarrow \quad & 2 \mathrm{X}^2+3 \mathrm{Y}^2-\mathrm{Z}^2=25\end{array}$
Asked in: AP EAMCET 2023 (18 May Shift 1)
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