Question

The angle between the pair of straight lines represented by the equation ${x^2} + \lambda xy + 2{y^2} + 3x - 5y + 2 = 0,$ is ${\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right)$ where $'\lambda '$ is a non-negative real number then $\lambda $ is
A. $2$
B. $0$
C. $3$
D. $1$

Answer 1

Hint: In order to solve this type of question, first we need to write the given equation and the general equation. Then, we have to compare both the equations to get the values of a, h and b. Next, we will use the formula for finding the angle between the lines and substitute the values in it to get the correct answer.

Formula used:
$a{x^2} + b{y^2} + 2hxy + 2gx + 2fy + c = 0$
$\tan \theta = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$

Complete step by step solution:
We are given the following pair of lines,
${x^2} + \lambda xy + 2{y^2} + 3x - 5y + 2 = 0$ ………………….equation$\left( 1 \right)$
We know that the general equation is,
$a{x^2} + b{y^2} + 2hxy + 2gx + 2fy + c = 0$ ………………….equation$\left( 2 \right)$
On comparing equation $\left( 1 \right)$ and $\left( 2 \right)$ we get,
$a = 1,\;b = 2,h = \dfrac{\lambda }{2},$
Finding the angle between the lines,
$\tan \theta = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
\[\theta = {\tan ^{ - 1}}\left( {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right)\]
We are given that,
$\theta = {\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right)$
Substituting the value of $\theta $, a, b and h in the above equation,
${\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{2\sqrt {{{\left( {\dfrac{\lambda }{2}} \right)}^2} - 1 \times 2} }}{{1 + 2}}} \right)$
${\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{2\sqrt {\dfrac{{{\lambda ^2}}}{4} - 2} }}{3}} \right)$
Solving it,
${\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right) = {\tan ^{ - 1}}\left[ {\dfrac{{2\left( {\dfrac{{\sqrt {{\lambda ^2} - 8} }}{2}} \right)}}{3}} \right]$
${\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{\sqrt {{\lambda ^2} - 8} }}{3}} \right)$
On comparing both the sides,
$\dfrac{1}{3} = \dfrac{{\sqrt {{\lambda ^2} - 8} }}{3}$
$\sqrt {{\lambda ^2} - 8} = 1$
Squaring both sides,
${\lambda ^2} - 8 = 1$
Solving for $\lambda ,$ we get,
$\lambda = 3$
$\therefore $ The correct option is C.

Note: To solve this question one has to be very careful while writing the general equation and comparing the values of $\theta ,$ a, h and b. Moreover, avoid any type of calculation mistakes while substituting and solving the equation to get the correct angle. While solving the value of $\lambda$,we get negative value as well but we need to neglect that value.