Senior two mathematics book 2 hyperbolic unit training questions and answers.
First, multiple-choice questions (6 points for each small question, ***42 points)
1. If the equation =- 1 represents a hyperbola whose focus is on the y axis, then the range of its half focal length c is ().
A.(0， 1) B.( 1，2) C.( 1，+？ None of the above is true.
Answer: c
Analysis: = 1, and the focus is on the y axis, then m-1> 0 and | m |-2 >；; 0, so m>2, c => 1.
2.(20 10 Jiangsu Nanjing first model, 8) If the distance from the focus of a hyperbola to the asymptote is equal to the length of the real axis, then the eccentricity e of this hyperbola is equal to ().
A.B. C. D。
Answer: c
Analysis: Let hyperbolic equation be = 1, then the distance from F(c, 0) to y= x is =2a b=2a, e= 2a, e =.
3.(20 10 Hubei key middle school simulation, 1 1) has the same asymptote as hyperbola = 1, and the hyperbolic equation passing through point (-3,4) is ().
A.= 1 B. = 1
C.= 1 D. = 1
A: A.
Analysis: Let hyperbola be =? = =- 1, so choose a.
4. Let the hyperbola c with eccentricity e: =1(a >; 0, b>0), the right focus is f, the straight line L passes through the point F, and the slope is k, then the necessary and sufficient condition for the intersection of the straight line L and the hyperbola C at the left and right branches is ().
a . k2-E2 & gt； 1 b . k2-E2 & lt； 1
E2-k2 & gt； 1d . E2-k2 & lt； 1
Answer: c
Analysis: What is the slope of the hyperbolic asymptote? , the straight line L intersects the left and right branches of the hyperbola, then-
5. The polygons in the figure below are all regular polygons, m and n are the midpoints of their sides, and hyperbolas are all concentrated on F 1 and F2 in the figure. Let the eccentricity of hyperbola in Figure ① ② ③ be e 1, e2 and e3 respectively, then ().
a . e 1 & gt； e2 & gtE3 b . e 1 & lt； e2 & lte3 p = " " & gt& lt/E2 & lt； e3 & gt
C.e 1=e3 e2
Answer: d
Analysis: e 1=+1,
For ②, if the side length of a square is 2, |MF2|=, |MF 1|= 1, |F 1F2|=2,
? E2 =；
Let |MF 1|= 1 for ③, then |MF2|=,? |F 1F2|=2，
? e3= + 1。
You pay one+1> , so e1= E3 > e2。
6.(20 10 Hubei provincial key middle school simulation, 1 1) It is known that the eccentricity of ellipse E is E, the two focuses are F 1 and F2, the parabola C takes F 1 as the vertex, F2 as the focus, and P is the intersection of two curves. If =e, the value of e is
A.B. C. D。
A: A.
Analysis: Let P(x0, y0), then ex0+a=e(x0+3c) e=.
7.(20 10 Jiangsu Nantong Nine Schools Simulation, 10) Known hyperbola =1(A > The right focus of 0, b>0) is f, the right directrix intersects an asymptote at point A, and the area of △OAF is (o is the origin), then the included angle between the two asymptotes is ().
.30 caliber? B.45？ C.60？ D.90？
Answer: d
Analysis: a (), S△OAF= c= a=b, so the two asymptotes are y=? X, the angle is 90? .
Two. Fill in the blanks (5 points for each small question, *** 15 points)
8. Known ellipse = 1, hyperbola =1(m >; 0, n>0) has the same focus F 1, F2, let the intersection of two curves be q,? QF 1F2=90？ , the eccentricity of hyperbola is _ _ _ _ _ _ _ _.
Answer:
Analysis: a2 = 25, b2= 16, c= =3.
| qf 1 |+qf2 | = 2a = 10，| qf2 |-| qf 1 | = 2m，
? |QF2|=5+m，|QF 1|=5-m。
| qf2 | 2 = | qf 1 | 2+| f 1 F2 | 2，
That is, (5+m)2=(5-m)2+62 m=,
? e= =。
9.(20 10 Hubei Huanggang No.1 model, 15) If a directrix of hyperbola = 1 happens to be the tangent of circle x2+y2+2x=0, then k is equal to _ _ _ _ _ _ _ _ _.
Answer: 48
Analysis: Because the circular equation is (x+ 1)2+y2= 1, it is -=-2, that is, =2, and k=48.
10. hyperbola-y2 =1(n >; The two focuses of 1) are f 1, F2, and p are on a hyperbola, and | pf1| pf2 | = 2, then the area of △PF 1F2 is _ _ _ _ _ _ _ _.
Answer: 1
Analysis: Let |PF 1| >|PF2|, then | pf1|-pf2 | = 2, so |PF 1|=, |PF2|=, and | f1F2 | 2 = △PF 1F2 is Rt△, so = |PF 1|? |PF2|= 1。
Third, answer questions (1 1? 13 is 10, 14 is 13, ***43)
1 1. If hyperbola =1(a > 0, b>0) is equal to the distance from the right focus to the left directrix, and the range of eccentricity e is found.
Analysis: As shown in the right figure, set point M(x0, y0) on the right branch of hyperbola. According to the meaning of the question, the distance from m point to right focus F2 is equal to its distance from left directrix |MN|, that is.
|MF2|=|MN|。
∫= e，？ =e，=e。
? x0=。
∵x0？ Answer.
∵ ? 1，e & gt 1,? E2-e & gt； 0.
? 1+e？ e2-e？ 1- ? e？ 1+ .
But e> 1,? 1
12. It is known that the area of △P 1OP2 is, and p is the bisector of the straight line P 1P2. Find a hyperbolic equation, taking the straight lines OP 1 and OP2 as asymptotes, passing through point P, and the eccentricity is.
Analysis: With O as the origin and the bisector of P 1OP2 as the X axis, a rectangular coordinate system as shown on the right is established, and the hyperbolic equation is =1(a > 0, b>0), determined by e2= = 1+( )2=( )2.
? The equations of the two asymptotes OP 1 and OP2 are y= x and y=- x respectively. Set points P 1(x 1, x 1) and point p2 (x2,-x2) (x1>; 0, x2>0), what about the ratio of point P? = =2. The coordinate of point P is (), that is, (), and point P is on the hyperbola = 1.
So = 1，
That is, (x1+2x2) 2-(x1-2x2) 2 = 9a2.
8x 1x2=9a2。 ①
|OP 1|= x 1，
|OP2|= x2，
sinP 1OP2=，
? = |OP 1|？ |OP2|？ sinP 1OP2=？ x 1x2？ = ,
X 1x2 =. ②.
A2= 4 from ① ②. b2=9，
So the hyperbolic equation is = 1.
13.(20 10 Jiangsu Yangzhou middle school simulation, 23) The known tilt angle is 45? The straight line L of passes through point A (1, -2) and point B, where b is in the first quadrant, and? |AB|=3。
(1) Find the coordinates of point B;
(2) If the straight line L and hyperbola C:-y2 =1(a >; 0) intersects at two different points, E and F, and the midpoint coordinate of line segment EF is (4, 1), so the value of number A is realistic.
Solution: (1) The straight AB equation is y=x-3, and the set point is B(x, y).
By and x>0, y>0, x=4, y= 1,? The coordinate of point B is (4, 1).
(2) by
(- 1)x2+6x- 10=0。
Let e (x 1, y 1) and f (x2, y2), then x 1+x2= =4 and a=2. At this time,? & gt0,? a=2。
14. As shown in the figure on the right, F 1 and F2 are the left and right focal points of hyperbola x2-y2= 1, the coordinates of point A are (,-), and point B is on hyperbola, but? =0.
(1) Find the coordinates of point B;
(2) verification:? F 1BA=？ F2BA。
(1) Analysis: According to the meaning of the question, we know that f1(-2,0), F2 (2,0),? A (,-).
Let B(x0, y0), then = (,-),? =(x0-，y0+)，
∵ ? =0,
? (x0- )- (y0+ )=0，
That is 3x0-y0=2.
∫x02-y02 = 1，
? x02-(3x0-2 )2= 1，
(2 x0-3)2=0。
? X0=, substitute 3x0-y0=2 to get y0=.
? The coordinates of point B are (,).
(2) Proof: = (-,-),? BF2=(，-)，=(-，-)，
cosF 1BA=，
cosF2BA=，