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  •  The equation of a pair of straight lines is <img alt="\mathrm{a x^2+2 h x y+b y^2=0.}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Ba%20x%5E2&plus;2%20h%20x%20ya

 The equation of a pair of straight lines is \mathrm{a x^2+2 h x y+b y^2=0.} By what angle must the axes be rotated so that the term containing \mathrm{xy} in the equation may be removed ?

 

Option: 1

\mathrm{\tan ^{-1} \frac{2 h}{a-b}}


Option: 2

\mathrm{\frac{1}{2} \tan ^{-1} \frac{2 h}{a-b}}


Option: 3

\mathrm{\frac{1}{2} \tan ^{-1} \frac{2 h}{a+b}}


Option: 4

\mathrm{\tan ^{-1} \frac{2 h}{a+b}}


Answers (1)

best_answer

Here the origin remains fixed. Let the axes be turned about the fixed origin through an angle \mathrm{\varphi } in the anticlockwise sense and new coordinates of the point \mathrm{\left ( x,y \right ) } become \mathrm{\left ( x',y' \right ) }.
Then the equations of transformation will be \mathrm{x=x^{\prime} \cos \varphi-y^{\prime} \sin \varphi }
\mathrm{\mathrm{y}=\mathrm{x}^{\prime} \sin \varphi+\mathrm{y}^{\prime} \cos \varphi\ \ \ \ \Rightarrow\ \ \therefore}  the changed equation will be 
\mathrm{\begin{aligned} & \mathrm{a}\left(\mathrm{x}^{\prime} \cos \varphi-\mathrm{y}^{\prime} \sin \varphi\right)^2+2 \mathrm{~h}\left(\mathrm{x}^{\prime} \cos \varphi-\mathrm{y}^{\prime} \sin \varphi\right)\left(\mathrm{x}^{\prime} \sin \varphi+\mathrm{y}^{\prime} \cos \varphi\right)+\mathrm{b}\left(\mathrm{x}^{\prime} \sin \varphi+\mathrm{y}^{\prime} \cos \varphi\right)^2=0 \\ & \begin{array}{r} \left(\mathrm{a} \cos ^2 \varphi+2 \mathrm{~h} \sin \varphi \cos \varphi+\mathrm{b} \sin ^2 \varphi\right) \mathrm{x}^{\prime 2}+(-\mathrm{a} \sin 2 \varphi+2 \mathrm{~h} \cos 2 \varphi+\mathrm{b} \sin 2 \varphi) \mathrm{x}^{\prime} \mathrm{y}^{\prime}+ \\ \\ \left(\mathrm{a} \sin ^2 \varphi-2 \sin \varphi \cos \varphi+\mathrm{b} \sin 2 \varphi\right) \mathrm{y}^{\prime 2}=0 \end{array} \end{aligned}}
the transformed equation of the pair of lines for the new axes will be 
\mathrm{(a \cos 2 \varphi+h \sin 2 \varphi+b \sin 2 \varphi) x^2+\{(b-a) \sin 2 \varphi+2 h \cos 2 \varphi\} x y+\left(a \sin ^2 \varphi-h \sin 2 \varphi+b \cos ^2 \varphi\right) y^2=0.}
This equation will not contain \mathrm{xy} if \mathrm{(b-a) \sin\ 2 \varphi+2 h\ \cos\ 2 \varphi=0}
or \mathrm{2 \varphi=\frac{2 h}{a-b} \quad \therefore \varphi=1 / 2 \tan ^{-1} \frac{2 h}{a-b} . \text { }} This is the required angle.
 

 

 

Posted by

Ritika Kankaria

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