A line segment AM = a moves in the XOY plane | vedclass

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

Question

(B) Let the coordinates of point $A$ be $(a \cos 	heta, a \sin 	heta)$ since it lies on the circle $x^2 + y^2 = a^2$.Since $AM$ is parallel to the $

Vedclass · Accepted Answer

$x^2 + y^2 = 2ax$

Vedclass · Answer

(B) Let the coordinates of point $A$ be $(a \cos 	heta, a \sin 	heta)$ since it lies on the circle $x^2 + y^2 = a^2$.Since $AM$ is parallel to the $X$-axis and has length $a$,the coordinates of $M$ can be represented as $(x, y) = (a \cos 	heta + a, a \sin 	heta)$ or $(a \cos 	heta - a, a \sin 	heta)$.Considering the case $(x, y) = (a \cos 	heta + a, a \sin 	heta)$,we have $x - a = a \cos 	heta$ and $y = a \sin 	heta$.Squaring and adding these equations:$(x - a)^2 + y^2 = a^2 \cos^2 	heta + a^2 \sin^2 	heta$$(x - a)^2 + y^2 = a^2$$x^2 - 2ax + a^2 + y^2 = a^2$$x^2 + y^2 = 2ax$.Similarly,for the other case,we get $x^2 + y^2 = -2ax$.

$A$ line segment $AM = a$ moves in the $XOY$ plane such that $AM$ is parallel to the $X$-axis. If $A$ moves along the circle $x^2 + y^2 = a^2$,then the locus of $M$ is

Similar Questions

The resultant capacitance between $A$ and $B$ in the figure is.....$\mu F$

Which of the following has the longest $C-O$ bond length? (Free $C-O$ bond length in $CO$ is $1.128 \ \mathring{A}$)

$A$ magnetic needle lying parallel to a magnetic field is turned through $60^{\circ}$. The work done on it is $W$. The torque required to maintain the magnetic needle in the position mentioned above is

$A$ real value of $x$ will satisfy the equation $\left( \frac{3 - 4ix}{3 + 4ix} \right) = \alpha - i\beta$ (where $\alpha, \beta$ are real),if

Which of the following is a homosporous pteridophyte?

Vedclass Test Series

Exam Paper Generator

Online Exam Module