Question 1 As shown in figure 1, it is known that the center of the ellipse is at the coordinate origin, the focus F 1F2 is on the X axis, the length of the long axis A 1A2 is 4, and the intersection of the left directrix L and the X axis is m, | ma1|: | a/kloc-.

(i) Find the equation of ellipse;

(2) if the straight line L 1: x = m (| m | > 1) and p is the moving point on L 1, mark the maximum point p of ∠F 1PF2 as q, and find the coordinates of point q (expressed in m).

1. Explore the source of the topic

1986 the fifth question of science and engineering mathematics in the national unified college entrance examination is as follows:

Question 2 has two points A(0, a) and B(0, b) in the positive direction of the Y axis, with B >；; A, try to find a point P in the positive direction of the X-axis, so as to maximize ∠APB, as shown in Figure 2.

Obviously, the second sub-topic of Zhejiang Juan Li 17 (text 19) in 2005 was obtained by changing the background of 1986 college entrance examination. The solutions to the two problems are exactly the same.

2. Examples of variants

In the review questions of the college entrance examination over the years, the questions adapted from the college entrance examination questions in 1986 are also very common. Here are two examples to illustrate.

Question 3: There is a section of the east-west highway L that ends at points A and B northwards. When observing the AB section from point P, the larger the ∠APB, the better the observation effect (as shown in Figure 3). Let |OA|=a, |OB|=b, (a>b). Where should the observation point P be located on the highway L in order to obtain the best observation effect?

Question 4: As shown in Figure 4, in the football match, Party A's winger dribbles the ball near Party B's goal and advances along the straight line L. How far is the winger's goal from Party B's bottom line and has the largest shooting angle? (Note: In Figure 4, AB stands for Party B's goal, the straight line where AB is located is Party B's bottom line, L stands for the advancing route of Party A's winger, and C stands for a certain position when Party A's winger advances. |AD|=a |。 b))。

Step 3 appreciate simple solutions

The above problems are generally solved by algebraic method, that is, by calculating the tangent value of the angle and establishing a functional relationship. In fact, from a geometric point of view, the solution will be simpler. Next we should work out the geometric solution of this problem.

Solution: consider the point p on l 1 above the x axis.

As shown in fig. 5, let the chord F 1F2 tangent to the point q l 1, and let p be any point different from q on l 1. Since the circumferential angle of the same arc is greater than the angle outside the circle, there is ∠ f1qf2 >: ∠ f/kloc.

According to secant theorem, there is qd2 = (m-1) (m+1) = m2-1,so QD=. So Q(m,).

According to symmetry, there is a point Q (m,-) below the X axis.

Note: If the distance from Y axis to l 1 is d, F2 is the center and D is the radius, and Y axis intersects with point O', then O' is the center and D is the radius, and the circle O' is the circle that cuts l 1 at point Q and passes through F 1 and F2.