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geometria analitica de lehmann - MATEMATICAS EJERCICIOS ...

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20<br />

22<br />

24<br />

1<br />

2<br />

3<br />

7<br />

8<br />

9<br />

11.<br />

12.<br />

13.<br />

4.<br />

5.<br />

7.<br />

9.<br />

12.<br />

14.<br />

15.<br />

20.<br />

21.<br />

24.<br />

SOLUCIONES A LOS <strong>EJERCICIOS</strong> 481<br />

2 25. 3x2 + 4y2 - 4x - 4 = 0.<br />

1 — sen 0 27. y2 = 4*3.<br />

eos (0 — co) = p. 28. i/2 + 8* - 16 = 0.<br />

2 + y2 - 4y = U. 29. (*2 + t/2 + 2 x ) 2 = 4(** + t/*).<br />

2 - 8x - 16 = 0. 30. (* 2 + t/2) 2 = 4 U 2 - t / 2) .<br />

T><br />

T ><br />

i v x<br />

f><br />

3 j i\<br />

2* 3 A<br />

(■• ?)•<br />

Grupo 39, p. 252<br />

4. (‘ VX f).<br />

( , ? ) . 5. ( V 5 , 35° 16'), ( V 6 , 324°44')<br />

o - í ) -<br />

?)• ( ! ^ T )'<br />

6.<br />

Polo.<br />

Polo. 10.<br />

Polo.<br />

«■ *)• (* í ) ' ( f f )■ ( f f)-<br />

1, (0,347,<br />

15.<br />

159°40').<br />

7,19.<br />

23. 0,966.<br />

6,46. 24. 2,35.<br />

r eos ^0 - = 4.<br />

Grupo 40, p. 259<br />

( V I , poIo><br />

(2 f)- (2- f ><br />

r eos (0 — co) = 1, en don<strong>de</strong> co = are tg (— %) está en el segundo cuadrante.<br />

r eos (0 — co) = 2, en don<strong>de</strong> co = are tg (%) está en el primer cuadrante,<br />

r eos 0 = — 3. 10. r sen 0 = 2.<br />

2r sen ( - y - 0^ + V 2 r sen ( 0 - -í.) = 4 V 2 sen i? .<br />

r2 - 12r eos (J j - 0^ + 20 = 0.<br />

r2 - 6r eos - 9 ^ + 6 VI - 4 = 0.<br />

Centro (2, 0), radio = 2. 22. Centro ^2, radio = 3.<br />

Centro^2, -y )' radio = 2. 23. Centro ( l, radio = 2 .<br />

r = — 2 eos 0; centro (1, it) , radio = 1.

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