Question

Difference of slopes of the lines represented by equation $x^2\left({\sec }^2\left(\theta \right)-{\sin }^2\left(\theta \right)\right)-2xy\tan \left(\theta \right)+y^2{\sin }^2\left(\theta \right)=0$ is

• A. $4$
• B. $3$
• C. $2$
• D. none of these

Answer 1

The correct option is A

$4$

Explanation for the correct option:

Step-1: Compare given equation with general equation.

Given, $x^2\left({\sec }^2\left(\theta \right)-{\sin }^2\left(\theta \right)\right)-2xy\tan \left(\theta \right)+y^2{\sin }^2\left(\theta \right)=0$

compare with general equation $a{x}^2+2hxy+b{y}^2=0$

$$\begin{gathered} a=\left({\sec }^2\left(\theta \right)-{\sin }^2\left(\theta \right)\right)......\left(i\right) \\ h=-\tan \left(\theta \right).......\left(ii\right) \\ b={\sin }^2\left(\theta \right).......\left(iii\right) \end{gathered}$$

Step-2: Apply condition for sum and product of slope.

We know that $m_1+m_2=-\frac{2h}{b}\&m_1·m_2=\frac{a}{b}$

From equation $\left(i\right),\left(ii\right)\&\left(iii\right)$.

$$\begin{gathered} \Rightarrow m_1+m_2=\frac{2\tan \left(\theta \right)}{{\sin }^2\left(\theta \right)} \\ \Rightarrow m_1+m_2=\frac{2}{\sin \left(\theta \right)\cos \left(\theta \right)}........\left(iv\right) \end{gathered}$$

and

$\Rightarrow m_1·m_2=\left(\frac{1}{{\cos }^2\left(\theta \right){\sin }^2\left(\theta \right)}-1\right).........\left(v\right)$

Step-3: Apply formula ${\left(a-b\right)}^2={\left(a+b\right)}^2-4ab$

$$\begin{gathered} \Rightarrow {\left(m_1-m_1\right)}^2={\left(m_1+m_2\right)}^2-4m_1{m}_2 \\ \Rightarrow {\left(m_1-m_1\right)}^2={\left(\frac{2}{\sin \left(\theta \right)\cos \left(\theta \right)}\right)}^2-4\left(\frac{1}{{\cos }^2\left(\theta \right){\sin }^2\left(\theta \right)}-1\right) \\ \Rightarrow {\left(m_1-m_1\right)}^2=\frac{4}{{\sin }^2\left(\theta \right){\cos }^2\left(\theta \right)}-\frac{4}{{\sin }^2\left(\theta \right){\cos }^2\left(\theta \right)}+4 \\ \Rightarrow {\left(m_1-m_1\right)}^2=4 \\ \Rightarrow \left(m_1-m_1\right)=2 \end{gathered}$$

Hence, correct answer is option $\left(A\right)$