College Trigonometry, 2011a

11.6 Hooked on Conics Again 987
11.6.4 Answers
1. x 2 +2xy + y 2 − x √ 2+y √ 2 − 6=0
becomes (x ′ ) 2 = −(y ′ − 3) after rotating
counter-clockwise through θ = π 4 .
y
2. 7x 2 − 4xy √ 3+3y 2 − 2x − 2y √ 3 − 5=0
becomes (x′ −2) 2
9
+(y ′ ) 2 = 1 after rotating
counter-clockwise through θ = π 3
y x ′
y ′ x ′
θ = π 4
y ′ θ = π 3
x
x
x 2 +2xy + y 2 − x √ 2+y √ 2 − 6=0
3. 5x 2 +6xy +5y 2 − 4 √ 2x +4 √ 2y =0
becomes (x ′ ) 2 + (y′ +2) 2
4
= 1 after rotating
counter-clockwise through θ = π 4 .
y
7x 2 − 4xy √ 3+3y 2 − 2x − 2y √ 3 − 5=0
4. x 2 +2 √ 3xy +3y 2 +2 √ 3x − 2y − 16 = 0
becomes(x ′ ) 2 = y ′ + 4 after rotating
counter-clockwise through θ = π 3
y x ′
y ′ x ′
θ = π 4
y ′ θ = π 3
x
x
5x 2 +6xy +5y 2 − 4 √ 2x +4 √ 2y =0
x 2 +2 √ 3xy +3y 2 +2 √ 3x − 2y − 16 = 0