Example 1: x2/9 - y2/16 = 1 Some textbooks and teachers switch the position of a and b in these equations. Follow the equation closely so you understand what’s going on. If you just memorize the equations you won’t be prepared when you see a different notation.
Example 1: x2/9 - y2/16 = 0
We’ll end up with an equation in the form (__ ± )( ± __) = 0. The first two terms need to multiply together to make x2/9, so take the square root and write it in those spaces: (x/3 ± __)(x/3 ± __) = 0 Similarly, take the square root of y2/16 and place it in the two remaining spaces: (x/3 ± y/4)(x/3 ± y/4) = 0 Since there are no other terms, write one plus sign and one minus sign so the other terms cancel when multiplied: (x/3 + y/4)(x/3 - y/4) = 0
We’ll end up with an equation in the form (__ ± )( ± __) = 0. The first two terms need to multiply together to make x2/9, so take the square root and write it in those spaces: (x/3 ± __)(x/3 ± __) = 0 Similarly, take the square root of y2/16 and place it in the two remaining spaces: (x/3 ± y/4)(x/3 ± y/4) = 0 Since there are no other terms, write one plus sign and one minus sign so the other terms cancel when multiplied: (x/3 + y/4)(x/3 - y/4) = 0
Example 1: Since (x/3 + y/4)(x/3 - y/4) = 0, we know x/3 + y/4 = 0 and x/3 - y/4 = 0 Rewrite x/3 + y/4 = 0 → y/4 = - x/3 → y = - 4x/3 Rewrite x/3 - y/4 = 0 → - y/4 = - x/3 → y = 4x/3
Example 2: (x - 3)2/4 - (y + 1)2/25 = 1 Set this equal to 0 and factor to get: ((x - 3)/2 + (y + 1)/5)((x - 3)/2 - (y + 1)/5) = 0 Separate each factor and solve to find the equations of the asymptotes: (x - 3)/2 + (y + 1)/5 = 0 → y = -5/2x + 13/2 ((x - 3)/2 - (y + 1)/5) = 0 → y = 5/2x - 17/2
Example 3: (y + 2)2/16 - (x + 3)2/4 = 1 Add the x term to both sides, then multiply each side by 16: (y + 2)2 = 16(1 + (x + 3)2/4) Simplify: (y + 2)2 = 16 + 4(x + 3)2
√((y + 2)2) = √(16 + 4(x + 3)2) (y+2) = ± √(16 + 4(x + 3)2)
In the equation (y+2) = ± √(16 + 4(x + 3)2), as x approaches infinity, the 16 becomes irrelevant. (y+2) = approximately ± √(4(x + 3)2) for large values of x
y + 2 = ±√(4(x+3)^2) y + 2 = ±2(x+3) y + 2 = 2x + 6 and y + 2 = -2x - 6 y = 2x + 4 and y = -2x - 8
