Question

The angle between the pair of lines joining the origin to the points of intersection of the line $3x-y+2=0$ and the curve $x^2+2xy+3y^2+4x+8y-11=0$ is

• A. ${\cos }^{-1}\left(\frac{2\sqrt{2}}{3}\right)$
• B. ${\tan }^{-1}\left(\frac{3}{\sqrt{17}}\right)$
• C. ${\tan }^{-1}\left(\frac{2\sqrt{2}}{3}\right)$
• D. ${\cos }^{-1}\left(\frac{\sqrt{2}}{3}\right)$

Answer 1

Answer: (C)

${\tan }^{-1}\left(\frac{2\sqrt{2}}{3}\right)$

$y=3x+2$
$x^2+2x(3x+2)+3(3x+2{)}^2+4x+8(3x+2)-11=0$

$7x^2+4x+3(9x^2+12x+4)+28x+5=0$

$34x^2+68x+17=0$

$2x^2+4x+1=0$

$x=\frac{-4\pm \sqrt{16-8}}{4}$

$x=\frac{-2\pm \sqrt{2}}{2}=-1\pm \frac{1}{\sqrt{2}}$

$y=3x+2$

$y=-3\pm \frac{3}{\sqrt{2}}+2$

$y=-1-\frac{3}{\sqrt{2}}$ & $y=-1+\frac{3}{\sqrt{2}}$

$\frac{y}{x}=\frac{\frac{-(3+\sqrt{2})\times 2}{\sqrt{2}}}{(-2+\sqrt{2})}$

$y=\frac{(3+\sqrt{2})}{1+\sqrt{2}}x$

& $\frac{y}{x}=\frac{2\times \left(\frac{3-\sqrt{2}}{\sqrt{2}}\right)}{\sqrt{2}-2}$

$=\frac{3-\sqrt{2}}{1-\sqrt{2}}$

$tan\theta =\left|\frac{(\frac{3-\sqrt{2}}{1-\sqrt{2}}-\frac{3+\sqrt{2}}{1+\sqrt{2}})}{\left(1-\frac{7}{1}\right)}\right|$

$=\left|\frac{+4\sqrt{2}}{6}\right|$

$tan\theta =\frac{2\sqrt{2}}{3}$