Question

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

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

Answer 1

The correct option is A

0

Calculate the value of the required equation.

It is given that $AM:MB=b:a$.

So, $AM$ divides $AB$externally in the ratio $b:a$.

Therefore, $x=\frac{\left[b.a\cos \beta -ab\cos \alpha \right]}{b-a}.....\left(i\right)$

Similarly, $y=\frac{\left[ba\sin \beta -ab\sin \alpha \right]}{b-a}.....\left(ii\right)$

Now, divide the equation $\left(i\right)$ by $\left(ii\right)$, we get

$\begin{array}{rcl}\frac{x}{y}& =& \frac{\left[\cos \beta -\cos \alpha \right]}{\left[\sin \beta -\sin \alpha \right]}\\ \frac{x}{y}& =& \frac{\left[\sin {\displaystyle \frac{\alpha +\beta }{2}}\right]}{\left[\cos {\displaystyle \frac{\alpha +\beta }{2}}\right]}\\ x.\cos \left(\frac{\alpha +\beta }{2}\right)+y.\sin \left(\frac{\alpha +\beta }{2}\right)& =& 0\end{array}$

Hence, option $\left(A\right)$ is correct.