Q59 / 160 · WBJEE 2012 · Mathematics

The line joining $A(b \cos \alpha, b \sin \alpha)$ and $B(a \cos \beta, a \sin \beta),$ where $a \neq b,$ is produced to the point $M(x, y)$ so that $A M: M B=b: a$. Then, $x \cos \frac{\alpha+\beta}{2}+y \sin \frac{\alpha+\beta}{2}$ is equal to

  • Correct answer
    A
    0
  • B
    1
  • C
    -1
  • D
    $a^{2}+b^{2}$

Solution

Since, $A M: B M=b: a$ $\therefore M$ divides $A B$ extemally in the ratio $b: a$...(i) $\therefore \quad x=\frac{b \cdot a \cos \beta-a b \cos \alpha}{b-a}$ $y=\frac{b a \sin \beta-a b \sin \alpha}{b-a}$ (ii) Divide Eq. (i) by Eq, (ii), we get $\frac{x}{y}=\frac{\cos \beta-\cos \alpha}{\sin \beta-\sin \alpha}=\frac{2 \sin \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}}{-2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}}$ $\therefore \quad x \cos \frac{\alpha+\beta}{2}+y \sin \frac{\alpha+\beta}{2}=0$

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

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