High school mathematics collection unit test questions

Analytical geometry examination paper

fill-in-the-blank question

1. The inclination range of the straight line passing through this point is, and the value range is.

2. The equation of a straight line passing through this point and equal to the distance between two points is.

3. When the chord length of a straight line and a circle is known to be the shortest, the equation is.

4. Given that a point is a moving point on a straight line, the minimum value of is.

5. The elliptic equation that has the same focus as the ellipse 4 x 2+9 y 2 = 36 and passes through the point (-3,2) is _ _ _ _ _ _ _ _ _ _ _ _.

6. If there is a point P( 1,-1) on the ellipse, f is the right focus of the ellipse, and there is a point m on the ellipse that minimizes the value of |MP|+2|MF|, then the minimum value is _ _ _ _ _ _.

7. If it is known as a point on an ellipse, the value range of is _ _ _ _ _ _ _ _ _.

8. It is known that A and B are two points on the ellipse += 1, and F2 is the right focus of the ellipse. If |AF2|+|BF2|=a, the distance from the midpoint of AB to the left line of ellipse is, and the elliptic equation is _ _ _ _ _ _ _ _ _ _ _.

9. There are three different points A (X 1, Y 1), B (6) and C (X3, Y3) on a hyperbola, and the distance to the focus f (0,5) becomes arithmetic progression, so y 1+y3 is equal to _ _.

10. Given a hyperbola, the range of the angle formed by the asymptote and the real axis is _ _ _ _ _ _ _ _.

1 1. If the vertex of a parabola is at the origin, the axis of symmetry is the coordinate axis, and the focus is on a straight line, then the parabola equation is _ _ _ _ _ _ _ _ _ _ _ _.

It is known that the left and right focal points of an ellipse are. If there is a point on the ellipse, the eccentricity range of the ellipse is.

13. If the length of a line segment cut by two parallel lines is, the inclination angle of the line segment can be.

① ② ③ ④ ⑤

The serial number of the correct answer is. (Write the serial numbers of all the correct answers)

14. It is known that the distance from a point on a parabola to the parabola directrix is, and the distance from a straight line is, then the minimum value of is.

Second, answer the question.

The midpoint of the line segment cut by the sum of two parallel lines of a straight line is on the straight line, and the length of the line segment cut by two parallel lines is 0. Find the equation of this straight line.

16. It is known that the moving circle passes through the fixed point (2,0) and is tangent to the straight line X =-2.

(1) The equation for finding the locus c of the center of motion;

(2) Is there a straight line L that makes L cross point (0,2) and intersect with trajectory C at two points, P and Q, and satisfies = 0? If it exists, find the equation of the straight line L; If it does not exist, explain why.

17. Points A and B are the left and right ends of the long axis of the ellipse, point F is the right focus of the ellipse, and point P is on the ellipse, above the axis. (1) Find the coordinates of point P; (2) Let m be a point on the long axis AB of the ellipse, and the distance from m to the straight line AP is equal to finding the minimum value of the distance from point to point M on the ellipse. ..

18. ellipse > > intersects a straight line at two points, and here is the origin of coordinates.

( 1);

(2) If the eccentricity of the ellipse satisfies ≤≤≤≤, find the range of the long axis of the ellipse.

19. A changing straight line L intersects an ellipse += 1 at two points P and Q, where m is the moving point on L and satisfies the relation | MP || MQ | = 2. If the slope of the straight line L is always equal to 1 in the process of change, the trajectory equation of the moving point M is found to explain the shape of the curve.

It is known that the center of the ellipse is at the coordinate origin and the focus is on the axis. The maximum distance from the point on the ellipse to the focus is 3, and the minimum distance is 1.

(1) Find the standard equation of ellipse;

(2) If the straight line intersects the ellipse at two points (not the left and right vertices), and the figure with diameter passes through the right vertex of the ellipse, verify that the straight line passes through a fixed point, and find the coordinates of the fixed point.

(-2,4) 2, x=3 or 4x+y- 13=0 3, 2x-y-5=0 4,1x-2y-1= 0 5,6,3 7, [

12、 13、①⑤ 14、

Part of the answer to the question

16, analysis: (1) As shown in the figure, let m be the center of the motion circle, f (2,0), the intersection point m be the perpendicular of the straight line x =-2, and the vertical foot be n,

From the meaning of the question: | MF | = | Mn |, that is, the distance from the moving point M to the fixed point F is equal to the alignment X =-2. From the definition of parabola, the trajectory of point M is a parabola, in which f (2,0) is the focus and X =-2 is the directrix.

Therefore, the equation of the center locus c of the moving circle is y2 = 8x.

(2) The equation of the straight line L can be set as x = k (y-2) (k ≠ 0).

From, y2-8ky+ 16k = 0,

δ=(-8k)2-4× 16k & gt； 0, and k < 0 or k> 1.

Let P(x 1, y 1) and Q(x2, y2), then y 1+y2 = 8k, y 1y2 = 16k,

From = 0，x 1x2+y 1y2 = 0，

That is, k2 (y1-2) (y2-2)+y1y2 = 0,

After finishing: (k2+1) y1y2-2k2 (y1+y2)+4k2 = 0,

The substitution result is16k (k2+1)-2k2.8k+4k2 = 0.

That is, 16k+4k2 = 0,

The solution is k =-4 or k = 0 (truncation),

So the straight line L exists, and its equation is x+4y-8 = 0.

17, solution: (1) omitted

(2) The equation of straight line AP is -+6 = 0. Set point m (,0), then the distance from m to straight line AP is.

So =, and -6≤≤≤6, then the solution is =2. Let the distance from the point (,) on the ellipse to the point m.

Since -6≤6, ∴ when =, d takes the minimum value.

18, [analysis]: let, by OP ⊥ OQ x 1 x 2+y 1 y 2 = 0.

Will again.

Substitute ① for simplification.

(2) Also known as (1)

, ∴ long axis 2a∑[].

19, [analysis]: let the moving point M(x, y) and the moving straight line l: y = x+m, let P(x 1, y 1) and Q(x2, y2) be the solutions of the equations, and eliminate y to get 3x2+4mx+2. 0,∴-<； m & lt，and x 1+x2 =-，x 1x2=，and ∵| MP | = | x-x 1 |，| MQ | = | x-x2 |。 From |MP||MQ|=2, we get | x-. So there are ∵ m = y-x, ∴| x2+2y2-4 | = 3. An ellipse with x2+2y2-4 = 3 is sandwiched between two arcs of a straight line and does not contain endpoints. From x2+2y2-4 =-3, an ellipse with x2+2y2 = 1

20, solution, let m∈R, in the plane rectangular coordinate system, known vector a = (MX, y+ 1), vector b=(x, y- 1), a⊥b,

The locus of the moving point M(x, y) is e.

(i) Find the equation of trajectory e and explain the curve shape represented by this equation;

(ii) It is well known. It is proved that there is a circle whose center is at the origin, so that any tangent of the circle and trajectory E always has two intersections A and B, OA⊥OB(O is the coordinate origin), and the equation of the circle is solved.

Solution: (1) Because a⊥b,

So a b = 0, that is, (MX, y+ 1) (x, y- 1) = 0,

So mx2+y2- 1=0, that is, mx2+y2= 1.

When m=0, the equation represents two straight lines;

When m = 1, the equation represents a circle;

When m > 0 and m≠ 1, the equation represents an ellipse;

When m < 0, the equation represents a hyperbola.

(2) At that time, the equation of trajectory E was set as a circle, and the equation was x2+y2 = R2 (0 < r < 1

When the tangent slope exists, the arbitrary tangent equation of the circle can be set to y=kx+t,

A(x 1，y 1)，B(x2，y2)，

therefore

That is t2=r2( 1+k2). ①