Post on 05-Aug-2021
MAT01191 – Vetores e Geometria Analıtica – Professora Miriam TelicheveskyLista de Exercıcios 12 – Gabarito
1. (a)(x− 8)2
4− (y + 5)2
36= 1
(b)(x+ 3)2
36+
(y − 4)2
9= 1.
(c) x2 = y − 7.
2. (a) x = x− 4, y = y + 3/2, a conica tem equacao x2
20− y2
25= 1.
(b) x = x− 3, y = y − 1, a conica tem equacao x2 = 5y.
(c) x = x− 2, y = y − 1, e a conica tem equacao x2 + 2y2 = 7.
3. (a) {x′ = 1
2x+
√32y
y′ = −√32x+ 1
2y
(b) {x′ =
√22x+
√22y
y′ = −√22x+
√22y
{x′ = −
√32x+ 1
2y
y′ = −12x−
√32y
4. (a) 2(3x2 + 2√
3xy+ y2) + 9(x2− 2√
3xy+ 3y2)− 72 = 0 ⇐⇒ 15x2− 14√
3xy+ 29y2− 72 = 0.
(b) 25(x2 − 2xy + y2)− 16(x2 + 2xy + y2)− 800 = 0 ⇐⇒ 9x2 − 72xy + 9y2 − 800 = 0.
(c) ERRATA DA QUESTAO: DEVERIA SER x′2 = −16y′, θ = π/2.
Resposta: y2 = 16x
5. (a)
(b)
(c)
6.
7. IMPORTANTE!!! Ao obter cos(2θ1) e tan(2θ1) e possıvel descobrir seu quadrante. Para quetenhamos certeza que θ1 e o menor angulo de rotacao possıvel e preciso que 2θ1 esteja no I ou IIquadrante, ou seja, cos e tan devem ter o mesmo sinal. Isso e impossıvel de decidir no item (b)pois cos se anula e tangente nao existe. Neste caso, vale que θ1 = 45o.
(a) i. A′ = 28 e C ′ = −8.
ii. tan 2θ1 = −√
3 e cos 2θ1 = −12.
iii. cos θ1 = 12
e sen θ1 =√32.
iv. {x = 1
2x′ −
√32y′
y =√32x′ + 1
2y′
v.x′2
2− y′2
7= 1.
(b) i. A′ = 4 e C ′ = 6
ii. tan 2θ1 nao existe e cos 2θ1 = 0.
iii. cos θ1 =√22
e sen θ1 =√22.
iv. {x =
√22x′ −
√22y′
y =√22x′ +
√22y′
v.x′2
3+y′2
2= 1.
(c) i. A′ = 5 e C ′ = −5.
ii. tan 2θ1 = 43
e cos 2θ1 = 35
iii. cos θ1 = 2√5
e sen θ1 = 1√5.
iv. {x = 2√
5x′ − 1√
5y′
y = 1√5x′ + 2√
5y′
v. x′2 − y′2 = 1.
(d) i.
ii. A′ = 169 e C ′ = 0.
iii. tan 2θ1 = −120/119 e cos 2θ1 = −119/169.
iv. cos θ1 = 513
e sen θ1 = 1213.
v. {x = 5
13x′ − 12
13y′
y = 1213x′ + 5
13y′
vi. x′2 =y′
13.
8. (a) i. A′ = −8 e C ′ = 28.
ii. tan 2θ2 = −√
3 e cos 2θ2 = 12.
iii. cos θ2 = −√32
e sen θ2 = 12.
iv. {x = −
√32x′ − 1
2y′
y = 12x′ −
√32y′
v.y′2
2− x′2
7= 1.
(b) i. A′ = 6 e C ′ = 4
ii. tan 2θ2 nao existe e cos 2θ2 = 0.
iii. cos θ2 = −√22
e sen θ2 =√22.
iv. {x = −
√22x′ −
√22y′
y =√22x′ −
√22y′
v.y′2
3+x′2
2= 1.
(c) i. A′ = −5 e C ′ = 5.
ii. tan 2θ2 = 43
e cos 2θ2 = −35
iii. cos θ1 = − 1√5
e sen θ2 = 2√5.
iv. {x = − 1√
5x′ − 2√
5y′
y = 2√5x′ − 1√
5y′
v. y′2 − x′2 = 1.
(d) i. A′ = 0 e C ′ = 169.
ii. tan 2θ2 = −120/119 e cos 2θ2 = 119/169.
iii. cos θ2 = −1213
e sen θ2 = 513.
iv. {x = −12
13x′ − 5
13y′
y = 513x′ − 12
13y′
v. y′2 =x′
13.
9. (a)
(b)
(c)
(d)