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\f4. 4x2+3y216x+6y+27=0 (4x2 16x)+ (3y2 + 6y) = 27 4(X2 4X) + 3(y2 + 2y) = 27 4m2 4x+(2)2)+3(y2 +2y+(1)2)=27+16+3 4(x 2)2 + 3(y + 1)2 = 8 5 PH This expression will always result to ""3 '3 negative. something positive. Degenerate case of There exists no point in the plane with _, ellipse real value coordinates x and y that will satisfy this equation thus it has NO GRAPH. 3. 3x2 + y2 -18x + 27 = 0 3(x2 - 6x)+y2 =-27 3(x2 - 6x + (3)2) + y? =-27+27=0 3(x - 3) 2+y 2 The only point in the plane that will satisfy the equation is (3 ,0). C (3, 0) The equation represents a point ellipse. Its graph is the point C(3, 0) which is also the center of the ellipse.MTH101A PRACTICE EXERCISE 5: CONIC SECTIONS I. Identify the graph of each equation 1. x' = 10y 7. x' -4y+ 2x-11= 0 2. 6x' + 9y' -24x-54y +115=0 8. 3x' = 4y? -12x 3. 9x' -16y' -36x - 64y-172= 0 9. x= 1-y 4. 5x' - 2y = 0 10. x' + 3y- +6=0 5. 3x' + 3y'+18x-12y -46=0 6. x2 - y? = 0\f