Question

If $\theta$ is the angle between the lines represented by $x^2-3xy+{\lambda y}^2+3x-5y+2=0$, where $\lambda$ is a real number, then ${cosec}^2\theta$ equals :

• A. 9
• B. 3
• C. 10
• D. 100

Answer 1

The correct option is A 9

$\begin{array}{c}\text{Solution - The guren equation of pain of lines is}\\ x^2-3xy+\lambda y^2+3x-5y+2=0\\ \text{Standard form is given by-}\\ a{x}^2+b{y}^2+2hxy+2gx+2fy+c=0\\ \text{for pair of lines}\end{array}$

$\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}1& -3/2& 3/2\\ -3/2& \lambda & -5/2\\ 3/2& -5/2& 2\end{array}\right|=0$

$\begin{array}{cc}& \text{on solving we get}\\ \Rightarrow 1(2\lambda -\frac{25}{4})+\frac{3}{2}(-3+\frac{15}{4})+\frac{3}{2}(\frac{15}{4}-\frac{3\lambda }{2})=0& \\ \Rightarrow & 2\lambda -\frac{25}{4}-\frac{9}{2}+\frac{45}{8}+\frac{45}{8}-\frac{9}{4}\lambda =0\\ \Rightarrow & \lambda =2\end{array}$

$\begin{array}{cc}\text{Now,}& \left|\frac{2\sqrt{n^2-ab}}{a+b}\right|\\ & =\left|\frac{2\times \sqrt{\frac{9}{4}-2}}{3}\right|=\frac{1}{3}\\ \tan \theta & =\mid +{\cot }^2\theta \\ {\mathrm{cosec}}^2\theta & =1+\frac{1}{{\tan }^2\theta }\end{array}$

$\begin{array}{cc}{\mathrm{cosec}}^2\theta & =1+9\\ {\mathrm{cosec}}^2\theta & =10\\ \text{Hence,}\left(c\right)& \text{is the correct option.}\end{array}$