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  • The equations to the straight lines passing through the origin and making an angle  

The equations to the straight lines passing through the origin and making an angle \mathrm{\alpha}  with the straight line \mathrm{y+x=0} are given by \mathrm{x^2+\mathrm{kxy} \sec 2 \alpha+y^2=0}, where \mathrm{\mathrm{k}=}

Option: 1

4


Option: 2

3


Option: 3

2


Option: 4

1


Answers (1)

best_answer

Let \mathrm{y=m x} be the equation of the straight line passing through the origin and making an angle \mathrm{\alpha \: with \: y+x=0}. Now, slope of line \mathrm{y+x=0\: is \: -1 }.
\mathrm{ \therefore \tan \alpha= \pm \frac{m+1}{1-m} \text { or } \tan ^2 \alpha=\frac{(m+1)^2}{(m-1)^2} }[squaring]

or \mathrm{ m^2\left(1-\tan ^2 \alpha\right)+2 m\left(1+\tan ^2 \alpha\right)+\left(1-\tan ^2 \alpha\right)=0. }
This is a quadratic equation in \mathrm{ m }, therefore there will be two values of \mathrm{ m } and hence two straight lines which make an angle \mathrm{ \alpha\: with \: y+x=0 }. Let the two values of \mathrm{ m\: be \: m_1 \: and \: m_2 }, then

\mathrm{ m_1+m_2=-\frac{2\left(1+\tan ^2 \alpha\right)}{1-\tan ^2 \alpha}=-2 \sec 2 \alpha \text { and } m_1 m_2=\frac{1-\tan ^2 \alpha}{1-\tan ^2 \alpha}=1 . }

Now the joint equation of the straight line passing through the origin and making an angle \mathrm{ \alpha \: with\: y+x=0 } is

\mathrm{ \left(y-m_1 x\right)\left(y-m_2 x\right)=0 }

\mathrm{ \text { or } \quad y^2-x y\left(m_1+m_2\right)+m_1 m_2 x^2=0 \text { or } y^2+2 x y \sec 2 \alpha+x^2=0 . }

Hence option 3 is correct.



 

Posted by

himanshu.meshram

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