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Senior two mathematics book 2 hyperbolic unit training questions and answers.
Many students always complain that they are not good at math. Actually, it's because the test questions are not done well. Mathematics needs a lot of exercises to help students understand the knowledge points. The following is the related information of hyperbolic unit training questions and answers in the second volume of senior two mathematics for you to read.

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=,