If the line $x+2 y=k$ intersects the curve $x^2-x y+y^2+3 x+3 y-2=0$ at two points $A$ and $B$ and if $O$ is…
If the line $x+2 y=k$ intersects the curve $x^2-x y+y^2+3 x+3 y-2=0$ at two points $A$ and $B$ and if $O$ is the origin, then the condition for $\angle \mathrm{AOB}=90^{\circ}$ is
$k^2+k+1=0$
$k^2-2 k+10=0$
$2 k^2+9 k-10=0$
$3 k^2+8 k-1=0$
Solution
We have,
$
x^2-x y+y^2+3 x+3 y-2=0
$
and
$
x+2 y=k \Rightarrow \frac{x+2 y}{k}=1
$
By homogeneous of Eq. (i), we get
$
\begin{aligned}
x^2-x y+y^2+3 x & \left(\frac{x+2 y}{k}\right) \\
& +3 y\left(\frac{x+2 y}{k}\right)-2\left(\frac{x+2 y}{k}\right)^2=0
\end{aligned}
$
$
\begin{aligned}
\Rightarrow k^2 x^2-k^2 x y+k^2 y^2+3 k x^2+6 k x y+3 k x y+6 k y^2 & \\
& -2 x^2-8 x y-8 y^2=0 \\
\Rightarrow x^2\left(k^2+3 k-2\right)-\left(k^2-9 k+8\right) & \\
x y+\left(k^2+6 k-8\right) y^2 & =0
\end{aligned}
$
Since, $\angle A O B=90^{\circ}$
$
\begin{aligned}
& \therefore k^2+3 k-2+k^2+6 k-8=0 \\
& \Rightarrow 2 k^2+9 k-10=0
\end{aligned}
$