If the lines, joining the origin to the points of intersection of the curve $2x^2-2xy+3y^2+2x-y-1=0$ and the line $x+2 y=k$, are at right angles, then $k^2$ equals

Equation of line $x+2y=k$

Also  $\frac{x+2y}{k}=1 \dots \left(1\right)$

$2x^2-2xy+3y^2+2x-y-1=0 \dots \left(2\right)$

Homogenise equation $\left(2\right)$ with respect to $\left(1\right)$

$2x^2-2xy+3y^2+\left(2x-y\right)\left(1\right)-1{\left(1\right)}^2=0$

$2x^2-2xy+3y^2+\left(2x-y\right)\left(\frac{x+2y}{k}\right)-1{\left(\frac{x+2y}{k}\right)}^2=0$

$\left(2x^2-2xy+3y^2\right)k^2+\left(2x-y\right)\left(x+2y\right)k-\left(x^2+4y^2+4xy\right)=0$

$x^2\left(2k^2+2k-1\right)+xy\left(-2k^2+3k-4\right)+y^2\left(3k^2-2k-4\right)=0$

Given these lines are perpendicular then

Coefficient of $x^2$$+$ Coefficient of $y^2$$=0$

$2k^2+2k-1+3k^2-2k-4=0$

$5k^2-5=0$

$k^2=1$