x²-2x+4y²+8y+1 = 0 The object is to make it look like this: (x-h)² (y-k)² —————— + ——————— = 1 a² b² which is an ellipse that looks like this "ᆼ" like an egg on a table, or: (x-h)² (y-k)² —————— + ——————— = 1 b² a² which is an ellipse that looks like the number "0". We can tell because a² > b² We start with this: x²-2x+4y²+8y+1 = 0 We get the 1 on the other side as a -1: x²-2x+4y²+8y = -1 It's good that the terms are in the order they are. Sometimes we have to switch them so that the terms are in the order "x², x, y², y" as they are here. Write it this way: [x²-2x]+[4y²+8y] = -1 The coefficient of x² is 1 so we don't do anything yet to the first bracket. But the coefficient of y² is 4, so let's factor that out in the 2nd bracket. [x²-2x]+[4(y²+2y)] = -1 Now I'll dispense with the brackets and just have parentheses: (x²-2x)+4(y²+2y) = -1 Next we want to make those two binomials into trinomials. We skip some space after those binomials (x²-2x )+4(y²+2y ) = -1 so we can add a number in those two boxes to make those binomials into trinomials so they'll factor into squares of binomials. (x²-2x+)+4(y²+2y+) = -1 Now let's figure out what number goes in the first box. The coefficient of x is -2 so we take half of it, getting -1, then we square -1, getting (-1)² or 1, so we add 1 where the first box is, but we also have to add 1 to the other side of the equation, like this: [x²-2x+4]+4(y²+2y+) = -1+1 Now let's figure out what number goes in the second box. The coefficient of y is 2 so we take half of it, getting 1, then we square 1, getting 1² or 1, but wait! See the 4 in front of the second parentheses? If we put a 1 in that box, It will get multiplied by the 4 in front of the parentheses. In other words putting a 1 in that second box will in effect amount to the same as adding 4 times 1 or 4 to the left side, not just 1. So we have to add 4(1) to the right side to offset adding 1 inside that parentheses since it will be multiplied by the 4, so we have: (x²-2x+4)+4(y²+2y+1) = -1+1+4(1) Notice that what's in the first parentheses, x²-2x+4 factors as (x-2)(x-2) or (x-2)² Also notice that what's in the second parentheses y²+2y+1 factors as (y+1)(y+1) or (y+1)². So this (x²-2x+4)+4(y²+2y+1) = -1+1+4(1) after substituting their factorization for the parentheses and combining the terms on the right, we have (x-2)²+4(y+1)² = 4 Next we get a 1 on the right by dividing all three terms by 4: (x-2)² 4(y+1)² 4 —————— + ——————— = ——— 4 4 4 And that simplifies to: (x-2)² (y+1)² —————— + ——————— = 1 4 1 which is in the form: (x-h)² (y-k)² —————— + ——————— = 1 a² b² because a² = 4 and b² = 1 The center is (h,k) = (2,-1) Plot it:Since a² = 4, a = 2 Since b² = 1, b = 1 a = 2 is the semi-major axis's length, so draw the major axis 2a or 4 units long with the center as the midpoint. We also draw the minor axis 2a or 4 units long with the center as the midpoint: Now we can sketch in the ellipse: The vertices are the endpoints of the major axes, (0,-1) and (4,-1). The covertices are the endpoints of the minor axis, (2,0) and (2,-2). Edwin