Question

The difference of the slopes of the lines represented by $x^2({\tan }^2\theta +{\cos }^2\theta )+2xy\tan \theta +y^2{\sin }^2\theta =0$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

$\begin{array}{c}\text{solution:}{x}^2({\tan }^2\theta +{\cos }^2\theta )+2xy\tan \theta +y^2{\sin }^2\theta =0\\ \text{If is a quadratic in}y\text{so its roots will}\\ \text{represent lines of form}\\ y=m_{1}x+c\text{and}y=m_{2}x+c\text{.}\end{array}$

$\begin{array}{c}\text{difference of roots}=\left|\frac{\sqrt{D}}{a}\right|\\ =\frac{\sqrt{4x^2{\tan }^2\theta -4x^2{\mathrm{sen}}^2\theta ({\tan }^2\theta +{\cos }^2\theta )}{\sin }^2\theta }{\sin \theta }\\ =\frac{2x\sqrt{{\tan }^2\theta -{\sin }^2\theta {\tan }^2\theta -{\sin }^2\theta {\cos }^2\theta }}{{\sin }^2\theta }\\ =\frac{2x(\sqrt{{\sec }^2\theta -{\tan }^2\theta -{\cos }^2\theta }}{{\sin }^2\theta }\end{array}$

$=\frac{2x\sqrt{1-{\cos }^2\theta }}{\sin \theta }=2x$

$=\frac{2x\sin \theta }{\sin \theta }=2x={\mathrm{lm}}_{1}x-m_{2}x|=|m_1-m_2\mid x$

here $2x=2$

so difference of slopes $=2$

Answer : option (B)