Question

The angle between the lines joining the origin to the points of intersection of the line $y=3x+2$ with the curve $x^2+2xy+3y^2+4x+8y=11,$ is

• A. $\frac{1}{2}{\tan }^{-1}\left(\frac{2\sqrt{2}}{3}\right)$
• B. ${\tan }^{-1}\left(\frac{\sqrt{2}}{3}\right)$
• C. ${\tan }^{-1}\left(\frac{2\sqrt{2}}{3}\right)$
• D. ${\tan }^{-1}\left(\frac{\sqrt{3}}{2\sqrt{2}}\right)$

Answer 1

The correct option is C ${\tan }^{-1}\left(\frac{2\sqrt{2}}{3}\right)$

The equation of the given straight line is
$y=3x+2$
$\Rightarrow \frac{y-3x}{2}=1...\left(1\right)$
The equation of the given curve is
$x^2+2xy+3y^2+4x+8y-11=0...\left(2\right)$

Making the eqn. $\left(2\right)$ homogeneous of second degree in $x$ and $y$ with the help of eqn. $\left(1\right),$ we get
$x^2+2xy+3y^2+4x\left(\frac{y-3x}{2}\right)+8y\left(\frac{y-3x}{2}\right)-11{\left(\frac{y-3x}{2}\right)}^2=0$
$\Rightarrow 4x^2+8xy+12y^2+8xy-24x^2+16y^2-48xy-11y^2+66xy-99x^2=0$
$\Rightarrow 7x^2-2xy-y^2=0$

On comparing the equation with $a{x}^2+2hxy+b{y}^2=0,$ we get
$a=7,b=-1$ and $h=-1.$

Let $\theta$ be the required angle. Then,
$\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}$
$\Rightarrow \tan \theta =\frac{2\sqrt{1+7}}{7-1}=\frac{2\sqrt{2}}{3}$
$\Rightarrow \theta ={\tan }^{-1}\left(\frac{2\sqrt{2}}{3}\right)$