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 ${\tan }^{-1}\left({\displaystyle \frac{1}{3}}\right)$ where ‘$\lambda$’ is non-negative real number, then find ‘$\lambda$’.

Answer 1

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

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 $\tan \theta =\left|{\displaystyle \frac{2\sqrt{h^2-ab}}{a+b}}\right|$
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 $h={\displaystyle \frac{-3}{2}}$.
Also the angle between the pair of straight lines represented by the equation $$x^2-3xy+\lambda y^2+3x-5y+2=0$$ is ${\tan }^{-1}\left({\displaystyle \frac{1}{3}}\right)$.
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 ${\tan }^{-1}\left({\displaystyle \frac{1}{3}}\right)$.
Therefore, $\theta ={\tan }^{-1}\left({\displaystyle \frac{1}{3}}\right)$………….(iii)
Now consider, $\tan \theta =\left|{\displaystyle \frac{2\sqrt{h^2-ab}}{a+b}}\right|$ and
$\Rightarrow \theta ={\tan }^{-1}\left({\displaystyle \frac{1}{3}}\right)$
From equation $\tan \theta =\left|{\displaystyle \frac{2\sqrt{h^2-ab}}{a+b}}\right|$ we can find out the value of the angle $\theta$.
As $\tan \theta =\left|{\displaystyle \frac{2\sqrt{h^2-ab}}{a+b}}\right|$
$\Rightarrow \theta ={\tan }^{-1}\left|{\displaystyle \frac{2\sqrt{h^2-ab}}{a+b}}\right|$ ……………..(iv)
As L.H.S of equation (iii) and (iv) are equal therefore their R.H.S are also equal therefore,
$\Rightarrow {\tan }^{-1}\left({\displaystyle \frac{1}{3}}\right)={\tan }^{-1}\left|{\displaystyle \frac{2\sqrt{h^2-ab}}{a+b}}\right|$
Now substituting the values as $a=1,b=\lambda$ and $h={\displaystyle \frac{-3}{2}}$ we get,
$\Rightarrow {\tan }^{-1}\left({\displaystyle \frac{1}{3}}\right)={\tan }^{-1}\left|{\displaystyle \frac{2\sqrt{{\left({\displaystyle \frac{-3}{2}}\right)}^2-1\times \lambda }}{1+\lambda }}\right|$
We can cancel out ${\tan }^{-1}$ from both sides. So we are now left with the equation
$\Rightarrow {\displaystyle \frac{1}{3}}={\displaystyle \frac{2\sqrt{{\left({\displaystyle \frac{-3}{2}}\right)}^2-\lambda }}{1+\lambda }}$
Solving the numerator of R.H.S we get,
$\Rightarrow {\displaystyle \frac{1}{3}}={\displaystyle \frac{2\sqrt{{\displaystyle \frac{9-4\lambda }{4}}}}{1+\lambda }}$.
By cross multiplication we get,
$\Rightarrow {\displaystyle \frac{1+\lambda }{1}}={\displaystyle \frac{3\times 2\times \sqrt{{\displaystyle \frac{9-4\lambda }{4}}}}{1}}$
$\Rightarrow {\displaystyle \frac{1+\lambda }{6}}=\sqrt{{\displaystyle \frac{9-4\lambda }{4}}}$
Taking square on both sides we get,
$\Rightarrow ({\displaystyle \frac{1+\lambda }{6}}{)}^2=(\sqrt{{\displaystyle \frac{9-4\lambda }{4}}}{)}^2$
By applying the identity $(a+b{)}^2=a^2+b^2+2ab$ and on further solving we get,
$\Rightarrow {\displaystyle \frac{1+{\lambda }^2+2\lambda }{36}}={\displaystyle \frac{9-4\lambda }{4}}$
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 $\lambda ={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}$
$\Rightarrow \lambda ={\displaystyle \frac{-38-\sqrt{1764}}{2}}={\displaystyle \frac{-38-42}{2}}={\displaystyle \frac{-80}{2}}=-40$ or
$\Rightarrow \lambda ={\displaystyle \frac{-38+\sqrt{1920}}{2}}={\displaystyle \frac{-38+42}{2}}={\displaystyle \frac{4}{2}}=2$
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 $\tan \theta =\left|{\displaystyle \frac{2\sqrt{h^2-ab}}{a+b}}\right|$
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 $\lambda$.