Question

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 $AN:NB=b:a$, then

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

Answer 1

Answer: (C)

$\frac{AN}{NB}=\frac{b}{a}$
Coordinate of N will be
$x=\frac{m_1{x}_2-m_2{x}_1}{m_1-m_2},y=\frac{m_1{y}_2-m_2{y}_1}{m_1-m_2}$
So, $x=\frac{b\cdot a\cos \beta -a\cdot b\cos \alpha }{b-a}$
$\Rightarrow dy=\frac{b\cdot a\sin \beta -a\cdot b\sin \alpha }{b-a}$
$\Rightarrow b-a=\frac{ba\cos \beta -ab\cos \alpha }{x}=\frac{ab\sin \beta -ab\sin \alpha }{y}$
$\Rightarrow \frac{b-a}{ab}=y\cos \beta -y\cos \alpha =x\sin \beta -x\sin \alpha$
So, $2y\sin \frac{\beta +\alpha }{2}\cdot \sin \frac{\alpha -\beta }{2}=2x\cos \frac{\beta +\alpha }{2}\sin \frac{\beta -\alpha }{2}$
$\Rightarrow -y\sin \frac{\beta +\alpha }{2}\sin \frac{\beta -\alpha }{2}=x\cos \frac{\beta +\alpha }{2}\sin \frac{\beta -\alpha }{2}$
So, $x\cos \frac{\beta +\alpha }{2}+y\sin \frac{\beta +\alpha }{2}=0$