OC starts from (0,0) AB is a line segment that has a midpoint C(because OC is touching it at C) Slope of OC is Tan60.Coordinates of A is (SQRT3,1)Find the equation of locus of point B.
OC starts from (0,0) AB is a line segment that has a midpoint C(because OC is touching it at C) Slope of OC is Tan60.Coordinates of A is (SQRT3,1)Find the equation of locus of point B.
1. The point C is located on the straight line l:
$\displaystyle l:\vec c = \langle 1, \sqrt{3} \rangle \cdot t$
2. Let $\displaystyle \vec b = \langle b_1, b_2 \rangle$ denotes the stationary vector of point B. C is the midpoint of $\displaystyle \overline{AB}$.
Therefore:
$\displaystyle \displaystyle{\frac{\vec b + \vec a}2 = \vec c}$ with $\displaystyle \vec a = \langle \sqrt{3} , 1 \rangle$
3. That means:
$\displaystyle \left| \begin{array}{rcl}\frac{b_1+\sqrt{3}}2 &=& t \\ \frac{b_2+1}2 &=& \sqrt{3} \cdot t \end{array} \right.$ .... $\displaystyle \implies$ .... $\displaystyle \left| \begin{array}{rcl} b_1 &=& 2t - \sqrt{3} \\ b_2 &=& 2 \sqrt{3} \cdot t - 1 \end{array} \right.$
4. The stationary vector of point B is:
$\displaystyle \vec b = \langle 2t - \sqrt{3}, 2 \sqrt{3} \cdot t - 1 \rangle~\implies~ \vec b = 2 \langle 1, \sqrt{3} \rangle \cdot t - \langle \sqrt{3} , 1 \rangle$
This is the equation of a parallel of line c.
Many Thanks for your answer. I managed to solve it like this:
A(SQRT3,1) is given. B is (x,SQRT3) and C is C(A,B).I applied the
midpoint formula.
First of all I found A as 2x- SQRT3 and B=2SQRT3x-1.I replaced A in B.
B=(A+SQRT3)SQRT3-1.Then B=SQRT3A+2.If we replace A and B with X and Y,
it becomes y= SQRT3x+2
Vectors are a different subject of analytic geometry in Turkey. Analytic Geometry subjects are: Analytic geometry:Lines ,tangents,symmetry vs..,Analytic of the circle,Vectors,Parabolia,Hyperbolia and Elips. Thanks again for you help.