Question

The line joining $A(b\cos \alpha ,b\sin \alpha )$ and $B(a\cos \beta ,a\sin \beta )$, where $a\ne b$, is produced to the point $M(x,y)$ so that $AM:MB=b:a$. 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

The correct option is A $0$

Since, $AM:BM=b:a$
$\therefore M$ divides $AB$ externally in the ratio $b:a$

$\therefore x=\frac{b\cdot a\cos \beta -ab\cos \alpha }{b-a}$ ...... (i)

$y=\frac{ba\sin \beta -ab\sin \alpha }{b-a}$ ..... (ii)

Divide Eq. (i) by Eq. (ii), we get
$\frac{x}{y}=\frac{\cos \beta -\cos \alpha }{\sin \beta -\sin \alpha }=\frac{2\sin \frac{\alpha -\beta }{2}\sin \frac{\alpha -\beta }{2}}{-2\cos \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}}$

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