Thank you for reporting, we will resolve it shortly
Q.
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
TS EAMCET 2018
Solution:
$\frac{A N}{N B}=\frac{b}{a}$
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}$
$\Rightarrow \, d y=\frac{b \cdot a \sin \beta-a \cdot b \sin \alpha}{b-a}$
$\Rightarrow \,b-a=\frac{b a \cos \beta-a b \cos \alpha}{x}=\frac{a b \sin \beta-a b \sin \alpha}{y}$
$\Rightarrow \, \frac{b-a}{a b}=y \cos \beta-y \cos \alpha=x \sin \beta-x \sin \alpha$
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, $x \cos \frac{\beta+\alpha}{2}+y \sin \frac{\beta+\alpha}{2}=0$
