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  • Find the difference of the tangents of the angles which the lines  <img alt="\mathrm{\left(\tan ^2 \alpha+\cos ^2 \alpha\right) x^2-2 x y \tan \alpha+\sin ^2 \alpha y^2=0}" src="https://e

Find the difference of the tangents of the angles which the lines  \mathrm{\left(\tan ^2 \alpha+\cos ^2 \alpha\right) x^2-2 x y \tan \alpha+\sin ^2 \alpha y^2=0} make with the axis of \mathrm{x}.

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

\frac{5}{2}


Answers (1)

best_answer

The given equation is

\mathrm{\left(\tan ^2 \alpha+\cos ^2 \alpha\right) x^2-2 x y \tan \alpha+\sin ^2 \alpha y^2=0}.....(i)

\mathrm{ \therefore \mathrm{a}=\tan ^2 \alpha+\cos ^2 \alpha, \mathrm{h}=-\tan \alpha, \mathrm{b}=\sin ^2 \alpha}


If the lines represented by (i) are \mathrm{y}-\mathrm{m}_1 \mathrm{x}=0$ and $\mathrm{y}-\mathrm{m}_2 \mathrm{x}=0,

 then  \mathrm\mathrm{m}_1+\mathrm{m}_2=-\frac{2 \mathrm{~h}}{\mathrm{~b}}=\frac{2 \tan \alpha}{\sin ^2 \alpha}=\frac{2}{\sin \alpha \cos \alpha}, and \mathrm{m}_1 \mathrm{~m}_2=\frac{\mathrm{a}}{\mathrm{b}}=\frac{\tan ^2 \alpha+\cos ^2 \alpha}{\sin ^2 \alpha}

Now, if the two lines make angles  \mathrm{\theta and \varphi} respectively to the axis of \mathrm{x}  then

\mathrm{ m_1=\tan \theta} and \mathrm{ m_2=\tan \varphi \quad}

\mathrm{ \therefore(\tan \theta-\tan \varphi)^2=\left(m_1-m_2\right)^2=\left(m_1+m_2\right)^2-4 m_1 m_2 }
\mathrm{ =\frac{4}{\sin ^2 \alpha \cos ^2 \alpha}-\frac{4\left(\sin ^2 \alpha+\cos ^2 \alpha\right)}{\sin ^2 \alpha}=\frac{4}{\sin ^2 \alpha \cos ^2 \alpha}\left[1-\cos ^2 \alpha\left(\tan ^2 \alpha+\cos ^2 \alpha\right)\right] }

\mathrm{ =\frac{4}{\sin ^2 \alpha \cos ^2 \alpha}\left[1-\sin ^2 \alpha-\cos ^4 \alpha\right]=\frac{4}{\sin ^2 \alpha \cos ^2 \alpha} }

\mathrm{ \left(\cos ^2 \alpha-\cos ^4 \alpha\right)=\frac{4 \cos ^2 \alpha\left(1-\cos ^2 \alpha\right)}{\sin ^2 \alpha \cos ^2 \alpha}=4}

Hence  \mathrm{ \tan \theta-\tan \varphi=2 }.

Posted by

Ajit Kumar Dubey

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