The line joining A(b cos α, b sin α) and B(a cos β, a sin β), where a ≠ b, is produced to the point M(x, y) so that A M: M B=b: a. Then, x cos (α+β/2)+y sin (α+β/2) is equal to

Question

The line joining A(b cos α, b sin α) and B(a cos β, a sin β), where a ≠ b, is produced to the point M(x, y) so that A M: M B=b: a. Then, x cos (α+β/2)+y sin (α+β/2) is equal to (A) 0 (B) 1 (C) -1 (D) a2+b2

Tardigrade Admin · Accepted Answer

Since, A M: B M=b: a ∴ M divides A B externally in the ratio b: a ∴ x=(b ⋅ a cos β-a b cos α/b-a) ...(i) y=(b a sin β-a b sin α/b-a) ...(ii) Divide Eq. (i) by Eq. (ii), we get (x/y)=( cos β- cos α/ sin β- sin α) =(2 sin (α+β/2) sin (α-β/2)/-2 cos (α+β)2 sin (α-β)2 ∴ x cos (α+β/2)+y sin (α+β/2)=0

Q. The line joining $A (b cos α, b sin α)$ and $B (a cos β, a sin β)$ , where $a \neq = b$ , is produced to the point $M (x, y)$ so that $A M : MB = b : a$ . Then, $x cos \frac{α + β}{2} + y sin \frac{α + β}{2}$ is equal to

$0$

$1$

$- 1$

$a^{2} + b^{2}$

Q. The line joining A(bcosα,bsinα) and B(acosβ,asinβ), where a=b, is produced to the point M(x,y) so that AM:MB=b:a. Then, xcos2α+β​+ysin2α+β​ is equal to

0

1

−1

a2+b2

Q. The line joining $A (b cos α, b sin α)$ and $B (a cos β, a sin β)$ , where $a \neq = b$ , is produced to the point $M (x, y)$ so that $A M : MB = b : a$ . Then, $x cos \frac{α + β}{2} + y sin \frac{α + β}{2}$ is equal to

$0$

$1$

$- 1$

$a^{2} + b^{2}$