Question

The line $3x+6y=k$ intersect the curve $2x^2+2xy+3y^2=1$ at points $A$ and $B$. The circle on $AB$ as diameter passes through the origin. The possible value of $k$ is

• A. 3
• B. 4
• C. -4
• D. -3

Answer 1

Answer: (A), (D)

We have $\frac{3x+6y}{k}=1$....(1)
$2x^2+2xy+3y^2-1=0\dots ..(2)$
Now homogenising (2) with the help of (1), we get

$\Rightarrow 2x^2+2xy+3y^2-{\left(\frac{3x+6y}{k}\right)}^2=0$
$\Rightarrow k^2\left(2x^2+3x+3y^2\right)-(3x+6y{)}^2=0$
Now coefficient of $x^2+$ coefficient of $y^2=0$
$\Rightarrow \left(2k^2-9\right)+\left(3k^2-36\right)=0\Rightarrow 5k^2=45\Rightarrow k^2=9\Rightarrow k=3\text{ or }-3$