y+x^2=3y+x2=3
Solve for y:
y=3-x^2y=3−x2
Substitute yy into x^2+4y^2=36x2+4y2=36
x^2+4(3-x^2)^2=36x2+4(3−x2)2=36
Write as the product of two binomials.
x^2+4(3-x^2)(3-x^2)=36color(white)(aaa)x2+4(3−x2)(3−x2)=36aaa
x^2+4(9-6x^2+x^4)=36color(white)(aaa)x2+4(9−6x2+x4)=36aaaMultiply the binomials
x^2+36-24x^2+4x^4=36color(white)(aaa)x2+36−24x2+4x4=36aaaDistribute the 4
4x^4-23x^2=0color(white)(aaa)4x4−23x2=0aaaCombine like terms
x^2(4x^2-23)=0color(white)(aaa)x2(4x2−23)=0aaaFactor out an x^2x2
x^2=0x2=0 and 4x^2-23=0color(white)(aaa)4x2−23=0aaaSet each factor equal to zero
x^2=0x2=0 and 4x^2=234x2=23
x=0x=0 and x=+-sqrt(23)/2color(white)(aaa)x=±√232aaaSquare root each side.
Find the corresponding yy for each xx using y=3-x^2y=3−x2
y=3-0=3, and, y=3-23/4=-11/4y=3−0=3,and,y=3−234=−114
Hence, the solutions are, (1) x=0, y=3; (2 and 3) x=+-sqrt23/2, y=-11/4(1)x=0,y=3;(2and3)x=±√232,y=−114.
Note that there are three solutions, which means there are three points of intersection between the parabola y+x^2=3y+x2=3 and the ellipse x^2+4y^2=36x2+4y2=36. See the graph below.
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