The line joining A(b cos alpha, b sin alpha) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $A(b \cos \alpha, b \sin \alpha)$ and $B(a \cos \beta, a \sin \beta)$.Since $M(x, y)$ divides $AB$ externally in the ratio $b : a$, the coor

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Given $A(b \cos \alpha, b \sin \alpha)$ and $B(a \cos \beta, a \sin \beta)$.Since $M(x, y)$ divides $AB$ externally in the ratio $b : a$, the coordinates of $M$ are given by the section formula for external division:$x = \frac{b(a \cos \beta) - a(b \cos \alpha)}{b - a} = \frac{ab(\cos \beta - \cos \alpha)}{b - a}$$y = \frac{b(a \sin \beta) - a(b \sin \alpha)}{b - a} = \frac{ab(\sin \beta - \sin \alpha)}{b - a}$Now, consider the expression $E = x \cos \frac{\alpha+\beta}{2} + y \sin \frac{\alpha+\beta}{2}$.Substituting $x$ and $y$:$E = \frac{ab}{b-a} [(\cos \beta - \cos \alpha) \cos \frac{\alpha+\beta}{2} + (\sin \beta - \sin \alpha) \sin \frac{\alpha+\beta}{2}]$Using the identity $\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$ and $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$:$E = \frac{ab}{b-a} [(-2 \sin \frac{\beta+\alpha}{2} \sin \frac{\beta-\alpha}{2}) \cos \frac{\alpha+\beta}{2} + (2 \cos \frac{\beta+\alpha}{2} \sin \frac{\beta-\alpha}{2}) \sin \frac{\alpha+\beta}{2}]$$E = \frac{2ab \sin \frac{\beta-\alpha}{2}}{b-a} [-\sin \frac{\alpha+\beta}{2} \cos \frac{\alpha+\beta}{2} + \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha+\beta}{2}]$$E = \frac{2ab \sin \frac{\beta-\alpha}{2}}{b-a} [0] = 0$.

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

Similar Questions

If the line $2x - 3y + 4 = 0$ divides the line segment joining the points $A(-2, 3)$ and $B(3, -2)$ in the ratio $m:n$,then the point which divides $AB$ in the ratio $-4m:3n$ is

If the distance between the points $(x, 2)$ and $(3, 4)$ is $2$,find the value of $x$.

Three vertices of a parallelogram taken in order are $(-1, -6)$,$(2, -5)$ and $(7, 2)$. The fourth vertex is

The line $L \equiv 6x + 3y + k = 0$ divides the line segment joining the points $(3, 5)$ and $(4, 6)$ in the ratio $-5: 4$. If the point of intersection of the lines $L = 0$ and $x - y + 1 = 0$ is $P(g, h)$,then $h =$

What is the distance between the points $(a, 0)$ and $(0, a)$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module