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.
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.