The line joining Ab cos - b sin - Ba cos - - a sin - is produced to the point Mx-y such that A M - MB- b - a then x cos -2-y sin - - -2

The correct option is B 0 Given that AM:BM=b:a , AMBM=AB+BMBM= ba⇒ABBM=b aa ∴ B divides AM in the ratio (b a):a ⇒ acosβ=(b a ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Section Formula

The line join...

Question

The line joining $A (b c o s α, b s i n α), B (a c o s β, a s i n β)$ is produced to the point M(x,y) such that A M : MB- b : a then $x c o s (\frac{α + β}{2}) + y s i n (\frac{α + β}{2})$

- 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a^{2} + b^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $0$
Given that $A M : B M = b : a$ ,
$\frac{A M}{B M} = \frac{A B + B M}{B M} = \frac{b}{a} \Rightarrow \frac{A B}{B M} = \frac{b - a}{a}$

$∴ B$ divides $A M$ in the ratio $(b - a) : a$
$\Rightarrow a cos β = \frac{(b - a) x + a b cos α}{b}$
$\Rightarrow (b - a) x = a b (cos β - cos α) \to (1)$
Also, $a sin β = \frac{(b - a) y + a b sin α}{b}$
$\Rightarrow (b - a) y = a b (sin β - sin α) \to (2)$

$\frac{1}{2} \Rightarrow \frac{x}{y} = \frac{cos β - cos α}{sin β - sin α} = \frac{- 2 sin (\frac{α + β}{2}) sin (\frac{α - β}{2})}{2 cos (\frac{α + β}{2}) sin (\frac{α - β}{2})}$

$\Rightarrow \frac{x}{y} = \frac{- sin \frac{α + β}{2}}{cos \frac{α + β}{2}}$
$\Rightarrow x cos \frac{α + β}{2} + y sin \frac{α + β}{2} = 0$

So, Option B is the correct answer.

Suggest Corrections

Similar questions

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 : M B = b : a

. Then

x cos \frac{α + β}{2} + y sin \frac{α + β}{2}

is equal to

Q. The line joining A

(b cos α, b sin α)

and B

(a cos β, a sin β)

is produced to the point

M (x, y)

so that

A M : M B = b : a

, then

x cos \frac{α + β}{2} + y sin \frac{α + β}{2} =

Q. The line joining

A

(b cos α, b sin α)

and

B

(a cos β, a sin β)

is produced to the point

M

(x, y)

, so that

A M

and

B M

are in the ration

b : a

. Prove that

x + y tan (\frac{α + β}{2}) = 0

Q. If

sin β \neq cos β

and

x, y

and

z

satisfy the equations

x cos α - y sin α + z = cos β + 1

x sin α + y cos α + z = - sin β + 1

x cos (α + β) - y sin (α + β) + z = 2

then

x^{2} + y^{2} + z^{2}

equals

Q. The point of intersection of tangents to the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

at two points

the sum of whose eccentric angles is constant, is

(a \frac{c o s \frac{α + β}{2}}{c o s \frac{α - β}{2}}, b \frac{s i n \frac{α + β}{2}}{c o s \frac{α - β}{2}})

Join BYJU'S Learning Program

The line joining A(bcosα,bsinα),B(acosβ,asinβ) is produced to the point M(x,y) such that A M : MB- b : a then xcos(α+β2)+ysin(α+β2)

The line joining $A (b c o s α, b s i n α), B (a c o s β, a s i n β)$ is produced to the point M(x,y) such that A M : MB- b : a then $x c o s (\frac{α + β}{2}) + y s i n (\frac{α + β}{2})$