1. Tardigrade
  2. Question
  3. Mathematics
  4. The line 3 x+6 y=k intersect the curve 2 x2+2 x y+3 y2=1 at points A and B. The circle on A B as diameter passes through the origin. The possible value of k is

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

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

Straight Lines

Solution:

We have $\frac{3 x+6 y}{k}=1$....(1)
$2 x ^2+2 xy +3 y ^2-1=0 \ldots . .(2)$
Now homogenising (2) with the help of (1), we get
image
$\Rightarrow 2 x^2+2 x y+3 y^2-\left(\frac{3 x+6 y}{k}\right)^2=0$
$\Rightarrow k^2\left(2 x^2+3 x+3 y^2\right)-(3 x+6 y)^2=0$
Now coefficient of $x^2+$ coefficient of $y^2=0$
$\Rightarrow\left(2 k^2-9\right)+\left(3 k^2-36\right)=0 \Rightarrow 5 k^2=45 \Rightarrow k^2=9 \Rightarrow k=3 \text { or }-3$