Question 1048628
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Solve the system and graph the curves:

(y-2)^2 = 9(x+2) ;and
9x^2+4y^2+18x-16y = 0
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(y-2)^2 = 9(x+2),               (1)
9x^2 + 4y^2 + 18x - 16y = 0.    (2)

Open parentheses in the first equation. Then the system is equivalent to

y^2 - 4y + 4 - 9(x+2) = 0,
9x^2 + 4y^2 + 18x - 16y = 0,   or

     -  9x + y^2  -  4y = 14,   (3)
9x^2 + 18x + 4y^2 - 16y =  0.   (4)

Multiply equation (3) by -4 (both sides), and then add to equation (3). You will get

9x^2 + 54x = -54,  or

x^2 + 6x + 6 = 0.

Apply the quadratic formula to find the roots. You will get

{{{x[1,2]}}} = {{{(-6 +- sqrt(36 - 4*6))/2}}} = {{{(-6 +- sqrt(12))/2}}} = {{{-3 +- sqrt(3)}}}.


For {{{x[1]}}} = {{{-3 + sqrt(3)}}}, you have from (1)

(y-2)^2 = 9*(x+2) = 9*(-1 + sqrt(3)), and {{{y[1,2]}}} = {{{2 +- 3*sqrt(sqrt(3)-1)}}}. 


For {{{x[2]}}} = {{{-3 - sqrt(3)}}}, you have from (1)

(y-2)^2 = 9*(x+2) = 9*(-1 - sqrt(3)), and the solution to (y-2) does not exist, since the right side {{{-1 - sqrt(3)}}} is negative.


So, there are two solutions (and, correspondingly, two intersection points)

 
     x = {{{3 - sqrt(3)}}}  and  {{{y[1]}}} = {{{2 + 3*sqrt(sqrt(3)-1)}}},  {{{y[2]}}} = {{{2 - 3*sqrt(sqrt(3)-1)}}}. 


To graph the curves, notice that

y - 2 = +/- {{{3*sqrt(x+2)}}},                  ( from (1), and )
9*(x+1)^2 - 9 + 4*(y-2)^2 - 16 = 0.     ( from (2) )   or


y       = {{{2 +- 3*sqrt(x+2)}}},
(y-2)^2 = {{{25/4 - (9/4)*(x+2)^2}}}.


So the plot is
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{{{graph( 330, 330, -6.5, 5.5, -5.5, 7.5,
          2 + 3*sqrt(x+2), 2 - 3*sqrt(x+2), 2 + sqrt(25/4-(9/4)*(x+2)^2), 2 - sqrt(25/4-(9/4)*(x+2)^2)  
)}}}


Line (1) (red + green) and Line (2).