Question

If the angle between the pair of straight lines represented by the equation ${x^2} - 3xy + \lambda {y^2} + 3x - 5y + 2 = 0$ is ${\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right)$ where ‘$\lambda$’ is non-negative real number, then find ‘$\lambda$’.

Answer 1

The angle between the lines represented by $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$ is given by $\tan \theta = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
$\Rightarrow \theta = {\tan ^{ - 1}}\left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
The lines are parallel if the angle between them is zero. Thus, the lines are parallel if $\theta = 0$
$\Rightarrow \tan \theta = 0 \Rightarrow \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right| = 0$
$\Rightarrow {h^2} = ab$

Complete step by step answer:
Given equation of the pair of straight lines represented by ${x^2} - 3xy + \lambda {y^2} + 3x - 5y + 2 = 0$…….(i)
As we know that the angle between the lines represented by $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$ is given by $\tan \theta = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
Consider $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$ …………..(ii)
On comparing equation (i) and (ii) we get the values as $a = 1,b = \lambda$ and $h = \dfrac{{ - 3}}{2}$.
Also the angle between the pair of straight lines represented by the equation ${x^2} - 3xy + \lambda {y^2} + 3x - 5y + 2 = 0$ is ${\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right)$.
Let $\theta$ be the angle between the pair of straight lines represented by the equation
$\Rightarrow {x^2} - 3xy + \lambda {y^2} + 3x - 5y + 2 = 0$ is ${\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right)$.
Therefore, $\theta = {\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right)$………….(iii)
Now consider, $\tan \theta = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$ and
$\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right)$
From equation $\tan \theta = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$ we can find out the value of the angle $\theta$.
As $\tan \theta = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
$\Rightarrow \theta = {\tan ^{ - 1}}\left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$ ……………..(iv)
As L.H.S of equation (iii) and (iv) are equal therefore their R.H.S are also equal therefore,
${\Rightarrow \tan ^{ - 1}}\left( {\dfrac{1}{3}} \right) = {\tan ^{ - 1}}\left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
Now substituting the values as $a = 1,b = \lambda$ and $h = \dfrac{{ - 3}}{2}$ we get,
${\Rightarrow \tan ^{ - 1}}\left( {\dfrac{1}{3}} \right) = {\tan ^{ - 1}}\left| {\dfrac{{2\sqrt {{{\left( {\dfrac{{ - 3}}{2}} \right)}^2} - 1 \times \lambda } }}{{1 + \lambda }}} \right|$
We can cancel out ${\tan ^{ - 1}}$ from both sides. So we are now left with the equation
$\Rightarrow \dfrac{1}{3} = \dfrac{{2\sqrt {{{\left( {\dfrac{{ - 3}}{2}} \right)}^2} - \lambda } }}{{1 + \lambda }}$
Solving the numerator of R.H.S we get,
$\Rightarrow \dfrac{1}{3} = \dfrac{{2\sqrt {\dfrac{{9 - 4\lambda }}{4}} }}{{1 + \lambda }}$.
By cross multiplication we get,
$\Rightarrow \dfrac{{1 + \lambda }}{1} = \dfrac{{3 \times 2 \times \sqrt {\dfrac{{9 - 4\lambda }}{4}} }}{1}$
$\Rightarrow \dfrac{{1 + \lambda }}{6} = \sqrt {\dfrac{{9 - 4\lambda }}{4}}$
Taking square on both sides we get,
$\Rightarrow {(\dfrac{{1 + \lambda }}{6})^2} = {(\sqrt {\dfrac{{9 - 4\lambda }}{4}} )^2}$
By applying the identity ${(a + b)^2} = {a^2} + {b^2} + 2ab$ and on further solving we get,
$\Rightarrow \dfrac{{1 + {\lambda ^2} + 2\lambda }}{{36}} = \dfrac{{9 - 4\lambda }}{4}$
Again by cross multiplication and on further simplification we get,
$\Rightarrow (1 + {\lambda ^2} + 2\lambda ) = 9 \times (9 - 4\lambda )$
$\Rightarrow {\lambda ^2} + 38\lambda - 80 = 0$
This is a quadratic equation so it must be having two roots. That means we get two values of $\lambda$ satisfying the above equation.
We get values by applying the formula $\lambda = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
$\Rightarrow \lambda = \dfrac{{ - 38 - \sqrt {1764} }}{2} = \dfrac{{ - 38 - 42}}{2} = \dfrac{{ - 80}}{2} = - 40$ or
$\Rightarrow \lambda = \dfrac{{ - 38 + \sqrt {1920} }}{2} = \dfrac{{ - 38 + 42}}{2} = \dfrac{4}{2} = 2$
Hence the values of $\lambda$ are $- 40$ and 2.

Since $\lambda$ is a non-negative real number so the value of $\lambda = 2.$

Note:
The angle between the lines represented by $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$ is given by $\tan \theta = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
The lines represented by $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$ are parallel ${h^2} = ab$.
Since $\lambda$ is a non-negative real number so we will only consider the positive value of $\lambda$.