Question

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

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

Answer 1

Answer: (A)

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