Question

If the line $x+2y=k$ intersects the curve $x^2-xy+y^2+3x+3y-2=0$ at two points $A$ and $B$ and if $O$ is the origin, then the condition for $\angle AOB=90^{\circ }$ is

• A. $k^2+k+1=0$
• B. $k^2-2k+10=0$
• C. $2k^2+9k-10=0$
• D. $3k^2+8k-1=0$

Answer 1

Answer: (C)

We have,
$x^2-xy+y^2+3x+3y-2=0\dots$ (i)
and
$x+2y=k\Rightarrow \frac{x+2y}{k}=1$
By homogeneous of Eq. (i), we get
$x^2-xy+y^2+3x\left(\frac{x+2y}{k}\right)$
$+3y\left(\frac{x+2y}{k}\right)-2{\left(\frac{x+2y}{k}\right)}^2=0$
$\Rightarrow k^2{x}^2-k^{2}xy+k^2{y}^2+3k{x}^2+6kxy+3kxy+6k{y}^2$
$-2x^2-8xy-8y^2=0$
$\Rightarrow x^2\left(k^2+3k-2\right)-\left(k^2-9k+8\right)$
$xy+\left(k^2+6k-8\right)y^2=0$
Since, $\angle AOB=90^{\circ }$
$\therefore k^2+3k-2+k^2+6k-8=0$
$\Rightarrow 2k^2+9k-10=0$