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 $\cos e{c}^2\theta$ is equal to

• A. $3$
• B. $9$
• C. $10$
• D. $100$

Answer 1

The correct option is C

$10$

Explanation for the correct option:

Step 1: Find the value of $\lambda$.

A equation of pair of straight lines is given:

$x^2-3xy+\lambda y^2+3x-5y+2=0...1$

Since, the general equation of pair of straight lines is:

$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0...2$

On comparing equation $1$ and equation $2$ we get,

$a=1,h=-\frac{3}{2},b=\lambda ,g=\frac{3}{2},f=-\frac{5}{2}$ and $c=2$.

Since, the Condition for pair of straight lines is:

$abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$

Substitute the obtained values as follows:

$$\begin{gathered} 2\lambda +\frac{45}{4}-\frac{25}{4}-\frac{9\lambda }{4}-\frac{9}{2}=0 \\ \Rightarrow \frac{8\lambda -9\lambda }{4}+\frac{45-25-18}{4}=0 \\ \Rightarrow \frac{-\lambda }{4}+\frac{2}{4}=0 \\ \Rightarrow -\lambda +2=0 \\ \Rightarrow \lambda =2 \end{gathered}$$

Therefore, the value of $\lambda$ is $2$.

Step 2: Compute the value of $\cos e{c}^2\theta$.

Since the angle between the pair of straight lines is given by:

$$\begin{gathered} \tan \theta =\frac{2\sqrt{h^2-ab}}{a+b} \\ \Rightarrow \tan \theta =\frac{2\sqrt{{\left({\displaystyle \frac{-3}{2}}\right)}^2-1·\lambda }}{1+\lambda } \\ \Rightarrow \tan \theta =\frac{2\sqrt{{\displaystyle \frac{9}{4}}-2}}{1+2} \\ \Rightarrow \tan \theta =\frac{2\sqrt{{\displaystyle \frac{9-8}{4}}}}{3} \\ \Rightarrow \tan \theta =\frac{2\sqrt{{\displaystyle \frac{1}{4}}}}{3} \\ \Rightarrow \tan \theta =\frac{1}{3} \end{gathered}$$

Since, $cot\theta =\frac{1}{\tan \theta }$.

Therefore, $cot\theta =3$.

Also, $\cos e{c}^2\theta =1+co{t}^2\theta$.

Therefore,

$$\begin{gathered} \cos e{c}^2\theta =1+3^2 \\ \Rightarrow \cos e{c}^2\theta =10 \end{gathered}$$

Since, the value of $\cos e{c}^2\theta$ is $10$.

Hence, option $C$ is correct .