Question

If the angle between the lines joining origin and the point of intersection of the line $x-y=2$ and the curve $x^2-4xy+2y^2-2x+y=0$ is ${45}^{\circ }$ then find $k$.

Answer 1

substitute $y=mx$ in $x-y=2$

we get,

$x=\frac{2}{1-m},y=\frac{2m}{1-m}$

substitute these values in equation, $x^2-4xy+2y^2-2x+y+k=0$

${\left(\frac{2}{1-m}\right)}^2-4\frac{2}{1-m}\frac{2m}{1-m}+2{\left(\frac{2m}{1-m}\right)}^2-2\frac{2}{1-m}+\frac{2m}{1-m}+k=0$

solving the above equation, we get,

$(k+6)m^2-(2k+10)m+k=0$..........(1)

$m=\frac{k+5\pm \sqrt{4k+25}}{k+6}$

$m_1-m_2=\frac{2\sqrt{4k+25}}{k+6}$..........(2)

$m_1{m}_2=\frac{k}{k+6}$........(3)

since given,

lines making an angle of ${45}^{\circ }$

$\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=1$

$(m_1-m_2{)}^2=(1+m_1{m}_2{)}^2$

From (2) and (3)

${\left(\frac{2\sqrt{4k+25}}{k+6}\right)}^2={(1+\frac{k}{k+6})}^2$

solving the above equation we get

$k=-1\pm \sqrt{17}$