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)