Question

The difference of the slopes of the lines $x^2({\sec }^2\theta -{\sin }^2\theta )-\left(2\tan \theta \right)xy+y^2{\sin }^2\theta =0$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

According to the question..........

$\begin{array}{c}Let,m_1\&m_2\\ sumoftheslope:m_1+m_2=\frac{-2h}{b}----\left(i\right)\\ and,\\ productofslope:m_1.m_2=\frac{a}{b}-----\left(ii\right)\\ Here,\\ a{x}^2+2hxy+b{y}^2=\mathrm{0.........}(generalequofstraightline.)\\ cofficientof:\\ a={\sec }^2\theta -{\sin }^2\theta \\ h=-\tan \theta \\ b={\sin }^2\theta \\ Now,valueputinto\\ sumoftheslope:m_1+m_2=\frac{-2h}{b}----\left(i\right)\\ \Rightarrow m_1+m_2=\frac{-2(-tan\theta )}{{\sin }^2\theta }=\frac{2\sin \theta \times 2}{2{\sin }^2\theta .\cos \theta }=\frac{4}{2sin\theta \cos \theta }=\frac{4}{\sin 2\theta }\\ and,\\ productofslope:m_1+m_2=\frac{a}{b}-----\left(ii\right)\\ \Rightarrow m_1.m_2=\frac{{\sec }^2\theta -{\sin }^2\theta }{{\sin }^2\theta }=\frac{1}{{\sin }^2\theta .{\cos }^2\theta }-1[divideby4\\ =\frac{4}{4{\sin }^2\theta .{\cos }^2\theta }-1=\frac{4}{{\left(\sin 2\theta \right)}^2}-1\\ Now,finddifference:\\ (m_1-m_2{)}^2=(m_1+m_2{)}^2-4m_1.m_2\\ ={\left(\frac{4}{\sin 2\theta }\right)}^2-4\left(\frac{4}{\left({\sin }^{2}2\theta \right)}-1\right)\\ =\frac{16}{\left({\sin }^{2}2\theta \right)}-\frac{16}{\left({\sin }^{2}2\theta \right)}+4\\ \Rightarrow (m_1-m_2{)}^2=4\\ \Rightarrow (m_1-m_2)=+\sqrt{4}=2\\ \therefore thediffereceofslopeis2.\\ So,thatthecorrectoptionisB.\end{array}$