Question

The equation $x^2-3xy+\lambda y^2+3x-5y+2=0$ When $\lambda$ is a real number, represents a pair of straight lines.If $\theta$ is the angle between the lines, then $cose{c}^2\theta =$ [EAMCET 1992]

• A. 3
• B. 9
• C. 10
• D. 100

Answer 1

The correct option is C

10

The equation $x^2-3xy+\lambda y^2+3x-5y+2=0$ represents a pair of straight lines.

$\therefore 2\lambda +2(-\frac{5}{2})\left(\frac{3}{2}\right)(-\frac{3}{2})-\frac{25}{4}-\frac{9\lambda }{4}-\frac{18}{4}=0\Rightarrow \lambda =2$

If $\theta$ is the angle between the lines, then

$\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}=\frac{2\sqrt{\left(\frac{9}{4}\right)-2}}{1+2}=\frac{1}{3}$

$\Rightarrow cose{c}^2\theta =1+co{t}^2\theta =1+9=10$.