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  • At what point the axes be shifted without rotation so that the equation <img alt="\mathrm{a x^2+2 h x y+b y^2+2 g x+2 f y+c=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7

At what point the axes be shifted without rotation so that the equation \mathrm{a x^2+2 h x y+b y^2+2 g x+2 f y+c=0} does not contain terms in x, y and constant term ?

 

Option: 1

\mathrm{\left(\frac{h g-a f}{a b-h^2}, \frac{h f-b g}{a b-h^2}\right)}


Option: 2

\mathrm{\left(\frac{h f-b g}{a b+h^2}, \frac{h g-a f}{a b+h^2}\right)}


Option: 3

\mathrm{\left(\frac{h f-b g}{a b-h^2}, \frac{h g-a f}{a b-h^2}\right)}


Option: 4

\mathrm{\left(\frac{h g-a f}{a b+h^2}, \frac{h f-b g}{a b+h^2}\right)}


Answers (1)

best_answer

Let the origin be shifted to \mathrm{\left ( x_{1},y_{1} \right )} Then, \mathrm{ x=X+x_{1}}  and \mathrm{Y+y_{1}}
Substituting \mathrm{x=\mathrm{X}+x_1 \text { and } \mathrm{y}=\mathrm{Y}+\mathrm{y}_1 \text { in } a x^2+2 h x y+b y^2+2 g x+2 f y+c=0}, we get 
\mathrm{\begin{aligned} & a\left(\mathrm{X}+x_1\right)^2+2 \mathrm{~h}\left(\mathrm{X}+x_1\right)\left(\mathrm{Y}+y_1\right)+b\left(\mathrm{Y}+y_1\right)^2+2 g\left(\mathrm{X}+x_1\right)+2 f\left(\mathrm{Y}+y_1\right)+c=0 \\ & a \mathrm{X}^2+2 h \mathrm{XY}+b \mathrm{Y}^2+2 \mathrm{X}\left(a x_1+h y_1+g\right)+2 \mathrm{Y}\left(h x_1+b y_1+f\right)+a x_1^2+2 h x_1 y_1+b y_1^2+2 \mathrm{~g} \mathrm{x}_1+2 f y_1+c=0 \end{aligned}}
This equation will be free from the terms containing \mathrm{X,Y} and constant term, if
\mathrm{\begin{aligned} & a x_1+h y_1+g=0\ \ \ \ \ .......\left ( i \right )\\ & h x_1+b y_1+f=0\ \ \ \ \ \ .........\left ( ii \right ) \end{aligned}}
\mathrm{\text { and, } a x_1^2+2 h x_1 y_1+b y_1^2+2 g x_1+2 f y_1+c=0 \ \ \ \ \ \ .........\left ( iii \right )}
\mathrm{Now, a x_1^2+2 h x_1 y_1+b y_1^2+2 g x_1+2 f y_1+c=0 }
\mathrm{\Rightarrow \quad x_1\left(a x_1+h y_1+g\right)+y_1\left(h x_1+b y_1+g\right)+\left(g x_1+g y_1+c\right)=0}
\mathrm{\begin{array}{ccc} \Rightarrow & x_1 \times 0+y_1 \times 0+g x_1+f y_1+c=0\ \ \ & \text { [Using (i) and (ii)] } \\ \Rightarrow & g x_1+f y_1+c=0 & \ldots \ldots \ldots \text { (iv) } \end{array}}
Solving \mathrm{\left ( i \right )} and \mathrm{\left ( ii \right )} by cross-multiplication, we get \mathrm{x_1=\frac{h f-b g}{a b-h^2} \quad, \quad y_1=\frac{h g-a f}{a b-h^2}}
The origin must be shifted at \mathrm{\left(\frac{h f-b g}{a b-h^2}, \frac{h g-a f}{a b-h^2}\right)}

 

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