Question 1208374
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1. Solve the system and graph the curves: (x + 1) ^ 2 + 2 * (y - 4) ^ 2 = 12; y ^ 2 - 8y = 4x - 16 
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Starting equations are

    (x+1)^2 + 2*(y-4)^2 = 12,    (1)

    y^2 - 8y = 4x - 16.          (2)


Rewrite equation (2) this way

    y^2 - 8y + 16 = 4x,

    (y-4)^2 = 4x.                (3)


Now replace the term  2*(y-4)^2  in equation (1) by  2*(4x), based on equation (3).

You will get simple quadratic equation for single unknown x

    (x+1)^2 + 8x = 12.


Write it in the standard form and solve by factoring 

    (x+1)^2 + 8x - 12 = 0,

    x^2 + 2x + 1 + 8x - 12 = 0,

     x^2 + 10x - 11 = 0,

     (x+11)*(x-1) = 0.


The roots are  x= -11  and  x= 1.



For x= -11, the corresponding values of y are

    (y-4)^2 = 4*(-11)    (from equation (3)

      y-4 = +/- {{{sqrt(-44)}}},  y = {{{4 +- i*sqrt(44)}}}   <<<---===  complex values



For x= 1, the corresponding values of y are

    (y-4)^2 = 4*1    (from equation (3)

      y-4 = +/- 2,    

which implies

      y = 4 + 2 = 6  or  y = 4 - 2 = 2.


Thus the given system (1), (2) has two real solutions  (x.y) = (1,6)   and  (x,y) = (1,2).


<U>ANSWER</U>.  The given system (1), (2) has two real solutions  (x.y) = (1,6)   and  (x,y) = (1,2).

         Equation (1) describes an ellipse; equation (2) describes a parabola
         having the horizontal axis, and these two curves have two intersection points  (1,6)  and  (1,2).
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Solved.


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