Question #b0b5a

1 Answer
Nov 30, 2017

# sqrt3x^2-xy=0.#

Explanation:

Let # S : sqrt3x^2-4xy+sqrt3y^2=0.#

# :. sqrt3x^2-3xy-xy+sqrt3y^2=0.#

#:. sqrt3x(x-sqrt3y)-y(x-sqrt3y)=0.#

#:. (sqrt3x-y)(x-sqrt3y)=0.#

Thus, the individual lines #l_1 and l_2" of "S# are,

#l_1 : sqrt3x-y=0, and, l_2 : x-sqrt3y=0.#

Observe that, #O(0,0) in l_1 : y=sqrt3x=xtan(pi/3),# and, is

making an #/_" of "pi/3# with the #+ve# direction of the #X-# Axis.

So, if #S# is rotated anti-clockwise by an #/_" of "pi/6# about #O# , then, so

would be the effect on both #l_1 and l_2.#

This means that, after the said rotation, #l_1# will now make an

#/_" of "(pi/3+pi/6)=pi/2# with the #+ve# direction of the #X-# Axis.

In other words, it will become the #Y-#Axis, or, #x=0.#

Similarly, #l_2# becomes, # y=xtan(pi/6+pi/6)=sqrt3x.#

Hence, the combined eqn. of lines becomes

#x(sqrt3x-y)=0, i.e., sqrt3x^2-xy=0.#

Enjoy Maths.!