Question

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 $\angle AOB =90^{\circ}$ is

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

Answer 1

Answer: (C)

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