A moving line intersects the lines $x+y=0$ and $x-y=0$ at the points A, B respectively such that the area of the triangle with vertices $(0,0), A \& B$ has a constant area $C$. The locus of the mid-point $A B$ is given by the equation

• A

  $\left(x^{2}+y^{2}\right)^{2}=C^{2}$
• Correct answer
  B

  $\left(x^{2}-y^{2}\right)^{2}=C^{2}$
• C

  $(x+y)^{2}=C^{2}$
• D

  $(x-y)^{2}=C^{2}$

Solution

Let mid point of $\mathrm{AB}=(\mathrm{h}, \mathrm{k})$ Let, $A=(\alpha,-\alpha), B=(\beta, \beta)$ $\therefore \beta+\alpha=2 \mathrm{~h}, \quad \beta-\alpha=2 \mathrm{k}$ Area $\Delta \mathrm{AOB}=\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}=\frac{1}{2} \sqrt{2 \alpha^{2}} \sqrt{2 \beta^{2}}=|\alpha \beta|=\mathrm{c}$ $\begin{array}{l} \therefore \alpha^{2} \beta^{2}=\mathrm{c}^{2} \Rightarrow 16 \alpha^{2} \beta^{2}=16 \mathrm{c}^{2} \Rightarrow\left\{(\beta+\alpha)^{2}-(\beta-\alpha)^{2}\right\}^{2}=16 \mathrm{c}^{2} \\ \Rightarrow\left(4 \mathrm{~h}^{2}-4 \mathrm{k}^{2}\right)^{2}=16 \mathrm{c}^{2} \rightarrow \text { locus }:\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right)^{2}=\mathrm{c}^{2} \end{array}$

Hence, the correct answer is option: 'B'.