Prova 2 de Geometria Analítica Ufal unificado

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Geometria Analítica (Resolução) – Prova 2 – 24/04/2015 Carlos Alberto 
 
𝑸𝒖𝒆𝒔𝒕ã𝒐 𝟏. 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑒 𝑜 𝑐𝑜𝑛𝑗𝑢𝑛𝑡𝑜 ℓ 𝑑𝑜𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 𝑃 𝑑𝑜 𝑝𝑙𝑎𝑛𝑜 𝑞𝑢𝑒 𝑗𝑢𝑛𝑡𝑎𝑚𝑒𝑛𝑡𝑒 𝑐𝑜𝑚 
𝐴 = (−3,5) 𝑒 𝐵 = (3,5) 𝑓𝑜𝑟𝑚𝑎𝑚 𝑢𝑚 𝑡𝑟𝑖â𝑛𝑔𝑢𝑙𝑜 𝑑𝑒 𝑝𝑒𝑟í𝑚𝑒𝑡𝑟𝑜 16. 
∗ 𝐷𝑎𝑑𝑜 𝑢𝑚 𝑝𝑜𝑛𝑡𝑜 𝑃 = (𝑥, 𝑦), 𝑡𝑒𝑚𝑜𝑠 𝑑(𝐴, 𝑃) + 𝑑(𝐴, 𝐵) + 𝑑(𝑃, 𝐵) = 16 
 
√(𝑥 + 3)2 + (𝑦 − 5)2 + √(3 + 3)2 + (5 − 5)2 + √(𝑥 − 3)2 + (𝑦 − 5)2 = 16 
√(𝑥 + 3)2 + (𝑦 − 5)2 + 6 + √(𝑥 − 3)2 + (𝑦 − 5)2 = 16 
√(𝑥 + 3)2 + (𝑦 − 5)2 + √(𝑥 − 3)2 + (𝑦 − 5)2 = 10 
√(𝑥 + 3)2 + (𝑦 − 5)2 = 10 − √(𝑥 − 3)2 + (𝑦 − 5)2 
(𝑥 + 3)2 + (𝑦 − 5)2 = 100 + (𝑥 − 3)2 + (𝑦 − 5)2 − 20√(𝑥 − 3)2 + (𝑦 − 5)2 
𝑥2 + 6𝑥 + 9 = 100 + 𝑥2 − 6𝑥 + 9 − 20√(𝑥 − 3)2 + (𝑦 − 5)2 
12𝑥 − 100 = 20√(𝑥 − 3)2 + (𝑦 − 5)2 
3𝑥 − 25 = 5√(𝑥 − 3)2 + (𝑦 − 5)2 
(3𝑥 − 25)2 = 25[(𝑥 − 3)2 + (𝑦 − 5)2] 
9𝑥2 − 150𝑥 + 625 = 25𝑥2 − 150𝑥 + 225 + 25𝑦2 − 250𝑦 + 625 
16𝑥2 + 25𝑦2 − 250𝑦 + 225 = 0 
16𝑥2 + 25(𝑦2 − 10𝑦) + 225 = 0 
16𝑥2 + 25(𝑦 − 5)2 − 625 + 225 = 0 
16𝑥2 + 25(𝑦 − 5)2 = 400 
𝑥2
25
+
(𝑦 − 5)2
16
= 1 
 
∗ 𝑂 𝑐𝑜𝑛𝑗𝑢𝑛𝑡𝑜 ℓ é 𝑢𝑚𝑎 𝑒𝑙𝑖𝑝𝑠𝑒 𝑐𝑢𝑗𝑜 𝑐𝑒𝑛𝑡𝑟𝑜 é 𝐶 = (0,5), 𝑒𝑖𝑥𝑜 𝑚𝑎𝑖𝑜𝑟 𝑚𝑒𝑑𝑖𝑛𝑑𝑜 10 𝑒 
𝑒𝑖𝑥𝑜 𝑚𝑒𝑛𝑜𝑟 8. 𝐶𝑜𝑚 𝑜𝑠 𝑓𝑜𝑐𝑜𝑠 𝑙𝑜𝑐𝑎𝑙𝑖𝑧𝑎𝑑𝑜𝑠 𝑠𝑜𝑏𝑟𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑦 = 5, 𝑚𝑎𝑖𝑠 𝑝𝑟𝑒𝑐𝑖𝑠𝑎𝑚𝑒𝑛𝑡𝑒, 
𝐹1 = (3,5) 𝑒 𝐹2 = (−3,5). 𝑉é𝑟𝑡𝑖𝑐𝑒𝑠 𝑉1 = (5,5), 𝑉2 = (−5,5), 𝑉3 = (0,9) 𝑒 𝑉4 = (0,1). 
 
 
Geometria Analítica (Resolução) – Prova 2 – 24/04/2015 Carlos Alberto 
𝑸𝒖𝒆𝒔𝒕ã𝒐 𝟐. 𝑆𝑒𝑗𝑎 ℱ 𝑢𝑚𝑎 𝑓𝑎𝑚í𝑙𝑖𝑎 𝑑𝑒 𝑟𝑒𝑡𝑎𝑠 𝑟 𝑜𝑛𝑑𝑒 𝑐𝑎𝑑𝑎 𝑟𝑒𝑡𝑎 é 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑑𝑎 𝑝𝑜𝑟 
𝑢𝑚 𝑛ú𝑚𝑒𝑟𝑜 𝑟𝑒𝑎𝑙 𝑚 𝑎𝑡𝑟𝑎𝑣é𝑠 𝑑𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑟: 𝑦 = 𝑚𝑥 −
1+𝑚2
𝑚
. 
 
𝑎) 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑎 𝑐ô𝑛𝑖𝑐𝑎 𝑞𝑢𝑒 é 𝑙𝑢𝑔𝑎𝑟 𝑔𝑒𝑜𝑚é𝑡𝑟𝑖𝑐𝑜 𝑑𝑜𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 𝑑𝑜 𝑝𝑙𝑎𝑛𝑜 𝑠𝑜𝑏𝑟𝑒 𝑜𝑠 𝑞𝑢𝑎𝑖𝑠 
𝑝𝑎𝑠𝑠𝑎 𝑢𝑚𝑎 ú𝑛𝑖𝑐𝑎 𝑟𝑒𝑡𝑎 𝑑𝑒𝑠𝑡𝑎 𝑓𝑎𝑚í𝑙𝑖𝑎. 
 
𝑚𝑦 = 𝑚2𝑥 − 1 − 𝑚2 
𝑚𝑦 = 𝑚2(𝑥 − 1) − 1 
𝑚2(𝑥 − 1) − 𝑚𝑦 − 1 = 0 
∆= 𝑦2 − 4(𝑥 − 1)(−1) 
∆= 𝑦2 + 4𝑥 − 4 
 
∗ 𝑈𝑚𝑎 𝑣𝑒𝑧 𝑞𝑢𝑒, 𝑐𝑎𝑑𝑎 𝑟𝑒𝑡𝑎 𝑝𝑎𝑠𝑠𝑎 𝑝𝑜𝑟 𝑢𝑚 ú𝑛𝑖𝑐𝑜 𝑝𝑜𝑛𝑡𝑜 𝑑𝑒𝑠𝑠𝑎 𝑐ô𝑛𝑖𝑐𝑎, 𝑒𝑛𝑡ã𝑜 ∆= 0. 
 
𝑦2 + 4𝑥 − 4 = 0 
4𝑥 = −𝑦2 + 4 
𝑥 = −
𝑦2
4
+ 1 
(𝑃𝑎𝑟á𝑏𝑜𝑙𝑎!) 
 
𝑏) 𝑆𝑜𝑏𝑟𝑒 𝑒𝑠𝑠𝑒 𝑝𝑎𝑟á𝑏𝑜𝑙𝑎, 𝑡𝑒𝑚𝑜𝑠 𝑣é𝑟𝑡𝑖𝑐𝑒 𝑉 = (1,0). 
𝑓𝑜𝑐𝑜: 𝑦2 = 4𝑐𝑥 → 𝑥 =
1
4𝑐
𝑦2 = −
1
4
𝑦2 ∴ 𝑐 = −1 ; 𝑓𝑜𝑐𝑜 𝐹 = (−1,0) 
𝑟𝑒𝑡𝑎 𝑑𝑖𝑟𝑒𝑡𝑟𝑖𝑧 𝑥 = 3 
 
 
Geometria Analítica (Resolução) – Prova 2 – 24/04/2015 Carlos Alberto 
𝑸𝒖𝒆𝒔𝒕ã𝒐 𝟑. 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑒 𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑞𝑢𝑎𝑑𝑟á𝑡𝑖𝑐𝑎 𝑎𝑏𝑎𝑖𝑥𝑜 𝑒 𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑎 𝑎𝑜𝑠 𝑖𝑡𝑒𝑛𝑠 𝑎 
𝑠𝑒𝑔𝑢𝑖𝑟. 
𝑥2 − 10√3𝑥𝑦 + 11𝑦2 + 16 = 0 
𝑎)𝐼𝑑𝑒𝑛𝑡𝑖𝑓𝑖𝑞𝑢𝑒 𝑎 𝑐ô𝑛𝑖𝑐𝑎 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑎𝑑𝑎 𝑝𝑒𝑙𝑎 𝑒𝑞𝑢𝑎çã𝑜; 
 
𝐸𝑞𝑢𝑎çã𝑜 𝑄𝑢𝑎𝑑𝑟á𝑡𝑖𝑐𝑎: 𝐴𝑥2 + 𝐵𝑥𝑦 + 𝐶𝑦2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0 
 ∆= 4𝐴𝐶 − 𝐵2 → 44 − 300 = −256 ; ∆< 0 ⇒ 𝐻𝑖𝑝é𝑟𝑏𝑜𝑙𝑒! 
 
𝐴 𝑝𝑟𝑒𝑠𝑒𝑛ç𝑎 𝑑𝑜 𝑡𝑒𝑟𝑚𝑜 𝑥𝑦 𝑛𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑒𝑣𝑖𝑑𝑒𝑛𝑐𝑖𝑎 𝑞𝑢𝑒 𝑎 𝑐ô𝑛𝑖𝑐𝑎 𝑎𝑠𝑠𝑜𝑐𝑖𝑎𝑑𝑎 à 𝑒𝑙𝑎 𝑒𝑠𝑡á 
𝑛𝑜 𝑟𝑜𝑡𝑎𝑐𝑖𝑜𝑛𝑎𝑑𝑎 𝑒𝑚 𝑟𝑒𝑙𝑎çã𝑜 𝑎𝑜 𝑠𝑖𝑠𝑡𝑒𝑚𝑎 𝑑𝑒 𝑐𝑜𝑜𝑟𝑑𝑒𝑛𝑎𝑑𝑎𝑠 𝑋𝑂𝑌. 𝑃𝑎𝑟𝑎 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑎𝑟𝑚𝑜𝑠 
𝑒𝑠𝑠𝑎 𝑐ô𝑛𝑖𝑐𝑎 𝑛𝑜 𝑠𝑖𝑠𝑡𝑒𝑚𝑎 𝑋′𝑂𝑌′, 𝑝𝑎𝑟𝑎 𝑜 𝑞𝑢𝑎𝑙 𝑜 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑑𝑜 𝑡𝑒𝑟𝑚𝑜 𝑥′𝑦′é 𝑛𝑢𝑙𝑜, 
𝑑𝑒𝑣𝑒𝑚𝑜𝑠 𝑎𝑐ℎ𝑎𝑟 𝑎 𝑓𝑜𝑟𝑚𝑎 𝑟𝑒𝑑𝑢𝑧𝑖𝑑𝑎 (𝑐𝑎𝑛ô𝑛𝑖𝑐𝑎) 𝑑𝑒𝑠𝑠𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑛𝑒𝑠𝑠𝑒 𝑠𝑖𝑠𝑡𝑒𝑚𝑎. 
 
𝑏) 𝐸𝑠𝑏𝑜ç𝑒 𝑎 𝑐ô𝑛𝑖𝑐𝑎. 
𝑥 = 𝑥′ cos 𝜃 − 𝑦′ sen 𝜃 
 𝑦 = 𝑥′ sen 𝜃 + 𝑦′ cos 𝜃 
 
∗ 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖𝑛𝑑𝑜 𝑛𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑒 𝑎𝑔𝑟𝑢𝑝𝑎𝑛𝑑𝑜 𝑜𝑠 𝑡𝑒𝑟𝑚𝑜𝑠, 𝑜𝑏𝑡𝑒𝑚𝑜𝑠: 
∗ 𝐶𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑑𝑜 𝑡𝑒𝑟𝑚𝑜 𝑥′𝑦′: 
−2 sen 𝜃 cos 𝜃 − 10√3 cos2 𝜃 + 10√3 sen2 𝜃 + 22 sen 𝜃 cos 𝜃 
 
𝑁𝑜 𝑠𝑖𝑠𝑡𝑒𝑚𝑎 𝑋′𝑂𝑌′, 𝑒𝑠𝑠𝑒 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 é 𝑛𝑢𝑙𝑜. 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜 … 
 
−2 sen 𝜃 cos 𝜃 − 10√3 cos2 𝜃 + 10√3 sen2 𝜃 + 22 sen 𝜃 cos 𝜃 = 0 
20 sen 𝜃 cos 𝜃 − 10√3 cos2 𝜃 + 10√3 sen2 𝜃 = 0 
2 sen 𝜃 cos 𝜃 − √3 cos2 𝜃 + √3 sen2 𝜃 = 0 
2 tg 𝜃 − √3 + √3 tg2 𝜃 = 0 
tg 𝜃 =
−2 ± 4
2√3
→ tg 𝜃 =
1
√3
 ∴ 𝜃 =
𝜋
6
𝑟𝑎𝑑 
 
∗ 𝐶𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑑𝑒 (𝑥′)2: 
 
cos2 𝜃 − 10√3 sen 𝜃 cos 𝜃 + 11 sen2 𝜃 
(
√3
2
)
2
− 10√3 (
1
2
) (
√3
2
) + 11 (
1
2
)
2
=
3
4
−
30
4
+
11
4
= −4 
 
∗ 𝐶𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑑𝑒 (𝑦′)2: 
 
sen2 𝜃 + 10√3 sen 𝜃 cos 𝜃 + 11 cos2 𝜃 
(
1
2
)
2
+ 10√3 (
1
2
) (
√3
2
) + 11 (
√3
2
)
2
=
1
4
+
30
4
+
33
4
= 16 
Geometria Analítica (Resolução) – Prova 2 – 24/04/2015 Carlos Alberto 
∗ 𝐸𝑞𝑢𝑎çã𝑜 𝑑𝑎 𝑐ô𝑛𝑖𝑐𝑎 𝑛𝑜 𝑠𝑖𝑠𝑡𝑒𝑚𝑎 𝑋′𝑂𝑌′: 
 
−4𝑥′2 + 16𝑦′2 + 16 = 0 
4𝑥′2 − 16𝑦′2 = 16 
(𝑥′)2
4
− (𝑦′)2 = 1 
 
∗ 𝐴 𝑐ô𝑛𝑖𝑐𝑎 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑎𝑑𝑎 𝑛𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑎𝑑𝑎 𝑖𝑛𝑖𝑐𝑖𝑎𝑙𝑚𝑒𝑛𝑡𝑒 é 𝑢𝑚𝑎 ℎ𝑖𝑝é𝑟𝑏𝑜𝑙𝑒 
𝑟𝑜𝑡𝑎𝑐𝑖𝑜𝑛𝑎𝑑𝑎 𝑛𝑜 𝑠𝑖𝑠𝑡𝑒𝑚𝑎 𝑋𝑂𝑌 𝑑𝑒 𝑢𝑚 â𝑛𝑔𝑢𝑙𝑜 𝑑𝑒 30° 𝑛𝑜 𝑠𝑒𝑛𝑡𝑖𝑑𝑜 𝑎𝑛𝑡𝑖 − ℎ𝑜𝑟á𝑟𝑖𝑜. 
∗ 𝐸𝑠𝑠𝑎 𝑐ô𝑛𝑖𝑐𝑎 𝑝𝑜𝑠𝑠𝑢𝑖 𝑐𝑒𝑛𝑡𝑟𝑜 𝐶 = (0,0), 𝑒𝑖𝑥𝑜 𝑟𝑒𝑎𝑙 𝑚𝑒𝑑𝑖𝑛𝑑𝑜 4 𝑢𝑛𝑖𝑑𝑎𝑑𝑒𝑠 𝑒 𝑒𝑖𝑥𝑜 
𝑖𝑚𝑎𝑔𝑖𝑛á𝑟𝑖𝑜 𝑚𝑒𝑑𝑖𝑛𝑑𝑜 2 𝑢𝑛𝑖𝑑𝑎𝑑𝑒𝑠. 
 
𝑐2 = 𝑎2 + 𝑏2 → 𝑐2 = 4 + 1 → 𝑐2 = 5 ∴ 𝑐 = ±√5 𝑢. 
 
∗ 𝐸 𝑑𝑖𝑠𝑡â𝑛𝑐𝑖𝑎 𝑓𝑜𝑐𝑎𝑙 𝑖𝑔𝑢𝑎𝑙 à 2√5 ; 
∗ 𝑁𝑜 𝑠𝑖𝑠𝑡𝑒𝑚𝑎 𝑋𝑂𝑌 𝑜 𝑒𝑖𝑥𝑜 𝑟𝑒𝑎𝑙 𝑒𝑠𝑡á 𝑠𝑜𝑏𝑟𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑦 =
1
√3
𝑥 𝑒 𝑜 𝑒𝑖𝑥𝑜 
𝑖𝑚𝑎𝑔𝑖𝑛á𝑟𝑖𝑜 𝑠𝑜𝑏𝑟𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 à 𝑒𝑠𝑠𝑎, 𝑦 = −√3𝑥. 
 
 
 
𝐹𝑜𝑐𝑜𝑠 𝐹1 = (√15 2⁄ , √5 2⁄ ) 𝑒 𝐹2 = (− √15 2⁄ , − √5 2⁄ ) 
𝑉é𝑟𝑡𝑖𝑐𝑒𝑠 𝐴1 = (√3, 1) 𝑒 𝐴2 = (−√3, −1) 
 
 
 
Geometria Analítica (Resolução) – Prova 2 – 24/04/2015 Carlos Alberto 
𝑸𝒖𝒆𝒔𝒕ã𝒐 𝟒. 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑒 𝑎 ℎ𝑖𝑝é𝑟𝑏𝑜𝑙𝑒 ℋ 𝑑𝑒 𝑒𝑞𝑢𝑎çã𝑜 
𝑥2
𝑎2
−
𝑦2
𝑏2
= 1. 
 
𝑎)𝑈𝑚𝑎 𝑟𝑒𝑡𝑎 é 𝑑𝑖𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 à 𝑢𝑚𝑎 ℎ𝑖𝑝é𝑟𝑏𝑜𝑙𝑒 𝑠𝑒 𝑎 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑎 𝑒𝑚 𝑢𝑚 ú𝑛𝑖𝑐𝑜 𝑝𝑜𝑛𝑡𝑜 𝑒 
𝑛ã𝑜 é 𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑎 à 𝑛𝑒𝑛ℎ𝑢𝑚𝑎 𝑑𝑒 𝑠𝑢𝑎𝑠 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎𝑠. 𝑀𝑜𝑠𝑡𝑟𝑒 𝑞𝑢𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑟:
𝑥𝑥0
𝑎2
−
𝑦𝑦0
𝑏2
= 1 
é 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 à ℋ 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 (𝑥0, 𝑦0) ∈ ℋ. 
 
𝐴 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑎𝑠 𝑟𝑒𝑡𝑎𝑠 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎𝑠 𝑑𝑎 ℎ𝑖𝑝é𝑟𝑏𝑜𝑙𝑒 𝑎𝑐𝑖𝑚𝑎 é 𝑦 = ±
𝑏
𝑎
𝑥 
𝐴 𝑟𝑒𝑡𝑎 𝑟: 
𝑥𝑥𝑜
𝑎2
−
𝑦𝑦𝑜
𝑏2
= 1 , 𝑝𝑜𝑠𝑠𝑢𝑖 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑚 =
𝑏2
𝑎2
.
𝑥𝑜
𝑦𝑜
 ; 𝐷𝑒𝑠𝑠𝑎 𝑓𝑜𝑟𝑚𝑎, 
𝑐𝑜𝑚𝑜 𝑜𝑠 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒𝑠 𝑎𝑛𝑔𝑢𝑙𝑎𝑟𝑒𝑠 𝑠ã𝑜 𝑑𝑖𝑠𝑡𝑖𝑛𝑡𝑜𝑠 𝑒𝑛𝑡𝑟𝑒 𝑠𝑖, 𝑐𝑜𝑛𝑐𝑙𝑢í𝑚𝑜𝑠 𝑞𝑢𝑒 𝑒𝑠𝑠𝑎 𝑟𝑒𝑡𝑎 
𝑛ã𝑜 é 𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑎 à𝑠 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎𝑠! 
𝑥0
2
𝑎2
−
𝑦0
2
𝑏2
= 1 → 𝑏2𝑥0
2 − 𝑎2𝑦0
2 = 𝑎2𝑏2 
 
∗ 𝐷𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑡𝑒𝑚𝑜𝑠 𝑦 =
𝑏2
𝑦0
(
𝑥𝑥0
𝑎2
− 1) ; 
 
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖𝑛𝑑𝑜 𝑛𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑎 ℎ𝑖𝑝é𝑟𝑏𝑜𝑙𝑒, 𝑜𝑏𝑡𝑒𝑚𝑜𝑠: 
 
𝑥2
𝑎2
−
𝑏2
𝑦0
2 (
𝑥2𝑥0
2
𝑎4
−
2𝑥𝑥0
𝑎2
+ 1) = 1 
𝑥2 (
𝑎2𝑦0
2 − 𝑏2𝑥0
2
𝑎4𝑦0
2 ) + 𝑥 (
2𝑏2𝑥0
𝑎2𝑦0
2 ) −
𝑏2
𝑦0
2 − 1 = 0 
𝑥2(𝑎2𝑦0
2 − 𝑏2𝑥0
2) + 𝑥(2𝑎2𝑏2𝑥0) − 𝑎
4(𝑏2 + 𝑦0
2) = 0 
 
∆= 4𝑎4𝑏4𝑥0
2 + 4𝑎4(𝑏2 + 𝑦0
2)(𝑎2𝑦0
2 − 𝑏2𝑥0
2) 
∆= 4𝑎6𝑏2𝑦0
2 + 4𝑎6𝑦0
4 − 4𝑎4𝑏2𝑥0
2𝑦0
2 
∆= 4𝑎4𝑦0
2(𝑎2𝑏2 + 𝑎2𝑦0
2 − 𝑏2𝑥0
2) 
 
𝑥2(−𝑎2𝑏2)+ 𝑥(2𝑎2𝑏2𝑥0) − 𝑎
4(𝑏2 + 𝑦0
2) = 0 
 
∗ 𝑃𝑒𝑙𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑎 ℎ𝑖𝑝𝑒𝑟𝑏𝑜𝑙𝑒, 𝑜 𝑡𝑒𝑟𝑚𝑜 𝑒𝑚 𝑑𝑒𝑠𝑡𝑎𝑞𝑢𝑒 𝑣𝑎𝑙𝑒 𝑧𝑒𝑟𝑜‼! 
∆= 0 
𝑥 = −
𝑏
2𝑎
=
−2𝑎2𝑏2𝑥0
−2𝑎2𝑏2
= 𝑥0 
 
∗ 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑎 𝑟𝑒𝑡𝑎 𝑟 é 𝑢𝑚𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 à ℎ𝑖𝑝𝑒𝑟𝑏𝑜𝑙𝑒 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 (𝑥0, 𝑦0). 
 
 
Geometria Analítica (Resolução) – Prova 2 – 24/04/2015 Carlos Alberto 
𝑏) 𝐸𝑛𝑐𝑜𝑛𝑡𝑟𝑒 𝑜 𝑝𝑜𝑛𝑡𝑜 𝑠𝑜𝑏𝑟𝑒 𝑎 ℎ𝑖𝑝é𝑟𝑏𝑜𝑙𝑒 
𝑥2
9
−
𝑦2
16
= 1 𝑚𝑎𝑖𝑠 𝑝𝑟ó𝑥𝑖𝑚𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑠: 
𝑦 = 2𝑥 + 1 
 
∗ 𝑆𝑎𝑏𝑒𝑛𝑑𝑜 − 𝑠𝑒 𝑞𝑢𝑒 𝑎 𝑚𝑒𝑛𝑜𝑟 𝑑𝑖𝑠𝑡â𝑛𝑐𝑖𝑎 𝑑𝑒 𝑢𝑚 𝑝𝑜𝑛𝑡𝑜 à 𝑢𝑚𝑎 𝑟𝑒𝑡𝑎 é 𝑢𝑚 𝑠𝑒𝑔𝑚𝑒𝑛𝑡𝑜 
𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 à 𝑚𝑒𝑠𝑚𝑎. 
∗ 𝑇𝑒𝑛𝑑𝑜 𝑝𝑜𝑟 𝑏𝑎𝑠𝑒 𝑜 𝑖𝑡𝑒𝑚 𝑎𝑛𝑡𝑒𝑟𝑖𝑜𝑟, 𝑒𝑠𝑠𝑒 𝑝𝑜𝑛𝑡𝑜 𝑞𝑢𝑒 𝑝𝑟𝑜𝑐𝑢𝑟𝑎𝑚𝑜𝑠 𝑒𝑛𝑐𝑜𝑛𝑡𝑟𝑎 − 𝑠𝑒 𝑛𝑎 
𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 à ℎ𝑖𝑝é𝑟𝑏𝑜𝑙𝑒 𝑞𝑢𝑒 é 𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑎 à 𝑟𝑒𝑡𝑎 𝑠. 
∗ 𝐶𝑜𝑚 𝑒𝑠𝑠𝑎𝑠 𝑖𝑛𝑓𝑜𝑟𝑚𝑎çõ𝑒𝑠, 𝑡𝑒𝑚𝑜𝑠 … 
𝑦 =
𝑏2
𝑦0
(
𝑥𝑥0
𝑎2
− 1) 
𝑦 =
16𝑥0
9𝑦0
𝑥 −
16
𝑦0
 
∗ 𝐶𝑜𝑚𝑜 𝑒𝑠𝑠𝑎 𝑟𝑒𝑡𝑎 é 𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑎 à 𝑟𝑒𝑡𝑎 𝑠, 𝑒𝑛𝑡ã𝑜 
16𝑥0
9𝑦0
= 2 ∴ 𝑦0 =
8
9
𝑥0 
 
∗ 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖𝑛𝑑𝑜 𝑛𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑎 ℎ𝑖𝑝é𝑟𝑏𝑜𝑙𝑒, 𝑜𝑏𝑡𝑒𝑚𝑜𝑠: 
 
𝑥0
2
9
−
(
64
81 𝑥0
2)
16
= 1 
16𝑥0
2 −
64
9
𝑥0
2 = 144 
80
9
𝑥0
2 = 144 
𝑥0 = ±
36
4√5
= ±
9
√5
 𝑒 𝑦0 = ±
8
√5
 
 
𝑇𝑒𝑚𝑜𝑠 𝑜𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 𝑃1 = (
9
√5
,
8
√5
) 𝑒 𝑃2 = (−
9
√5
, −
8
√5
) . 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑛𝑑𝑜 𝑎 𝑑𝑖𝑠𝑡â𝑛𝑐𝑖𝑎 𝑑𝑜𝑠 
𝑝𝑜𝑛𝑡𝑜𝑠 à 𝑟𝑒𝑡𝑎 𝑠, 𝑜𝑏𝑡𝑒𝑚𝑜𝑠: 
 
𝑑(𝑃1, 𝑠) =
|
8
√5
−
18
√5
− 1|
√5
=
10
√5
+ 1
√5
= 2 +
1
√5
 
 
𝑑(𝑃2, 𝑠) =
|−
8
√5
+
18
√5
− 1|
√5
=
10
√5
− 1
√5
= 2 −
1
√5
 
 
∗ 𝐶𝑜𝑚𝑝𝑎𝑟𝑎𝑛𝑑𝑜 𝑜𝑠 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑑𝑜𝑠, 𝑐𝑜𝑛𝑐𝑙𝑢í𝑚𝑜𝑠 𝑞𝑢𝑒 𝑜 𝑝𝑜𝑛𝑡𝑜 𝑃2 é 𝑜 𝑝𝑜𝑛𝑡𝑜 
𝑞𝑢𝑒 𝑝𝑒𝑟𝑡𝑒𝑛𝑐𝑒 à ℎ𝑖𝑝é𝑟𝑏𝑜𝑙𝑒 𝑞𝑢𝑒 𝑒𝑠𝑡á 𝑚𝑎𝑖𝑠 𝑝𝑟ó𝑥𝑖𝑚𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑠: 𝑦 = 2𝑥 + 1.