Given:
#3x^2-y^2=3" "........................Equation(1)#
#(x-1)^2+y^2=4" "................Equation(2)#
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#color(blue)("Determine the values of "x)#
Consider #Eqn(1)#
Write as: #y^2=3x^2-3" "..........Equation(1_a)#
Using #Eqn(1_a)# substitute for #y^2" in "Eqn(2)# giving:
#(x-1)^2color(white)("dddddd")+color(white)("d.")(3x^2-3)=4#
#x^2-2x+1color(white)("ddd")+color(white)("ddd")3x^2-3=4#
#4x^2-2x-2=4#
#4x^2-2x-6=0#
Divide both sides by 2
#2x^2-x-3=0#
#(2x-3)(x+1)=0#
For #2x-3=0color(white)("d")-> color(white)("ddd")x=3/2#
For #x+1=0color(white)("d")->color(white)("ddd")x=-1#
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#color(blue)("Determine the values of "y)#
#color(brown)("Substitute the value of "x=-1" in "Eqn(1) ) #
#3x^2-y^2=3 color(white)("ddd")->color(white)("ddd")3(-1)^2-y^2=3#
#color(white)("ddddddddddddd")->color(white)("ddd")y=0color(green)(larr" matches the graph")#
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#color(brown)("Substitute the value of "x=+3/2" in "Eqn(1) ) #
#3x^2-y^2=3 color(white)("ddd")->color(white)("ddd")3(3/2)^2-y^2=3#
#color(white)("ddddddddddddd")->color(white)("ddddd")27/4color(white)("d")-y^2=12/4#
#color(white)("ddddddddddddd")->color(white)("ddd")y=+-sqrt(15)/2#
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#color(blue)("Putting it all together")#
#(-1,0) ; (3/2,-sqrt(15)/2); (3/2,+sqrt(15)/2)#
