If the line joining the points $A(b \cos \alpha, b \sin \alpha)$ and $B(a \cos \beta, a \sin \beta)$ is extended to the point $N(x, y)$ such that $A N: N B=b: a$, then

• A

  $x \cos \frac{\alpha-\beta}{2}+y \sin \frac{\alpha+\beta}{2}=0$
• B

  $x \cos \frac{\alpha-\beta}{2}+y \sin \frac{\alpha-\beta}{2}=0$
• Correct answer
  C

  $x \cos \frac{\alpha+\beta}{2}+y \sin \frac{\alpha+\beta}{2}=0$
• D

  $x \cos \frac{\alpha+\beta}{2}+y \sin \frac{\alpha-\beta}{2}=0$

Solution

Coordinate of $N$ will be $$ x=\frac{m_1 x_2-m_2 x_1}{m_1-m_2}, \quad y=\frac{m_1 y_2-m_2 y_1}{m_1-m_2} $$ So, $$ x=\frac{b \cdot a \cos \beta-a \cdot b \cos \alpha}{b-a}, $$ $\Rightarrow \quad y=\frac{b \cdot a \sin \beta-a \cdot b \sin \alpha}{b-a}$ $\Rightarrow b-a=\frac{b a \cos \beta-a b \cos \alpha}{x}=\frac{a b \sin \beta-a b \sin \alpha}{y}$ $\Rightarrow \frac{b-a}{a b}=y \cos \beta-y \cos \alpha=x \sin \beta-x \sin \alpha$ So, $2 y \sin \frac{\beta+\alpha}{2} \cdot \sin \frac{\alpha-\beta}{2}=2 x \cos \frac{\beta+\alpha}{2} \sin \frac{\beta-\alpha}{2}$ $\Rightarrow-y \sin \frac{\beta+\alpha}{2} \sin \frac{\beta-\alpha}{2}=x \cos \frac{\beta+\alpha}{2} \sin \frac{\beta-\alpha}{2}$ So, $x \cos \frac{\beta+\alpha}{2}+y \sin \frac{\beta+\alpha}{2}=0$

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