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  • Mixed term XY is to be removed from the general equation of second degree <img alt="\mathrm{a x^2+b y^2+2 h x y+2 g x+2 f y+c=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cm

Mixed term XY is to be removed from the general equation of second degree

\mathrm{a x^2+b y^2+2 h x y+2 g x+2 f y+c=0}, one should rotate the axes through an

angle

\mathrm{\theta} given by Tan \mathrm{2\theta} equal to

Option: 1

\mathrm{(\mathrm{a}-\mathrm{b}) / 2 \mathrm{~h}}


Option: 2

\mathrm{2 \mathrm{~h} /(\mathrm{a}+\mathrm{b})}


Option: 3

\mathrm{(a+b) / 2 h}


Option: 4

\mathrm{2 \mathrm{~h} /(\mathrm{a}-\mathrm{b})}


Answers (1)

best_answer

d

Let \mathrm{\left(x^{\prime}, y^{\prime}\right)}, be the coordinates of new axes, then, put

           \mathrm{x=x^{\prime} \cos \theta-y^{\prime} \sin \theta},

\mathrm{y=x^{\prime} \sin \theta+y^{\prime \prime} \cos \theta}, in the equation.

Then coefficient of  \mathrm{x^{\prime} y^{\prime}} in transformed equation=0, so,

\mathrm{2(b-a) \sin \theta \cos \theta+2 h \cos 2 \theta=0}

or  \mathrm{\tan 2 \theta=2 h /(a-b)}, which is given in (d).

 

 

Posted by

Suraj Bhandari

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