When to use the function method to end the most valuable question of the conic curve

The maximum value problem of conic curves is commonly solved by the following methods:

When the conditions and conclusions of the problem can clearly reflect the geometric characteristics and significance, the combination of numbers and shapes can be considered. Function range solving method: When the conditions and conclusions of the problem can reflect a clear functional relationship, the objective function can be established first, and then the maximum value of the function can be found. The fundamental inequalities of algebra are used, combined with parametric equations, and the boundedness of trigonometric functions is exploited.

Conic curves: Learn to pay attention to these points, the definition and corresponding parameters must be mastered. Some problems take a lot of time to die, and using definitions is almost instantaneous. It often appears in the most valuable questions Pay attention to some geometric relations.

In conic curve problems, knowledge of the properties of triangle centroids, similar triangles, and congruent plane geometry is often used. This often comes up in the trajectory category. Special attention is paid to the knowledge of the position relationship between straight lines and conic curves, and the college entrance examination rate in recent years is almost 100%.

Pay special attention to the design of the intersection without seeking. This piece of knowledge is often difficult, and the difficulty is not that it is unthinkable, but that it cannot be calculated. Therefore, it is necessary to strengthen computing power in normal times.

Frequently Asked Questions: Fixed Value, Parameter Range, Midpoint Chord, etc., After mastering the basics, you must learn some knowledge that cannot be taught in class, and dealing with some problems can achieve twice the result with half the effort. I recommend these few:

Polar coordinates, parametric equations, hard solution theorem for conic curves, derivative of implicit functions, poles and poles of conic curves. Polar coordinates can be described as a spike for the problems related to the over-focus straight line, and the parametric equations can be used for certain range problems. The hard solution theorem is available in 80% of the conic curve problems, but the formula is complicated, I pushed it myself several times at the time, and then used it every time, and after this familiarization, some common problems can be solved in 10 minutes.

The derivative of the implicit function and the pole pole of the conic curve have the same effect, both of which are used to solve the midpoint string problem, which is faster than the spread method. Note: Polar coordinates and hard solution theorems and parametric equations can be answered on the answer sheet.

Other cautious, big questions are honest spread method, small questions are used secretly. Hopefully, happy learning.

Suppose point 1 on the parabola is (a,4a).

distance d=|4a-4a²-5|(4 +1) is the smallest numerator.

4a²-4a+5

4a²-4a+1+4

2a-1)²+4

So when a=1 2, the minimum = 1

d=|4a²-4a+5|/√17

So a=1 2 is minimal.

4a = 1, so is (1, 2, 1).

On a parabola, the point where the tangent is the same as the slope of the straight line has the shortest distance, which is (,1).

Let m(x,y), i.e., (y x-1)*(y x+1)=my2=m(x2-1).

If the trajectory of x 2-(y2 m) = 1 point m is an ellipse focused on the x-axis (excluding points a and b), then the range of m is .

m is less than -1 If the trajectory of the point m is a hyperbola with an eccentricity of 2 (minus a, b), then the value of m is 3

With a vector, the inner product is 0

Depending on the conditions, it will be fine.

The coordinates of p, q are (x1, -x1-1), (x2, -x2-1).

then x1x2+x1x2+x1+x2+1=0....

a^2=2c^2,b^2=c^2

x1^2/2c^2+y1^2/c^2=1...1)

x2^2/2c^2+y2^2/c^2=1...2)

x1^2-x2^2)/2c^2+(y1+y2)(y1-y2)/c^2=0

x1^2-x2^2)/2c^2+(-x1-x2-2)(x2-x1)/c^2=0...3)

by (*)3).

x1-x2)(

i.e.: (x1-x2) (

by (*) and (4).

Apparently x1!=x2

x1x2=1/6,x1+x2=-4/3

x1 2+x2 2) 2c 2+(y1 2+y2 2) sail fluid c 2=2

x1+x2)^2-2x1x2)/2c^2+((x1-x2-2)^2-2(x1+1)(x2+1))/c^2=2

i.e.: 16 Sakura 9-1 3) 2c 2+((4 3-2) 2-2(1 6-4 3+1)) c 2=2

That is: 13 18c 2 + 14 18c 2 = 2

c^2=3/4

a^2=3/2,b^2=3/4

Ellipse x 2 (3 2) + y 2 spine sedan hu (3 4) = 1

In the first question, the value of pf is equal to the distance from p to a 2 divided by two (the line) multiplied by e, the minimum is, the maximum is not very clear, it should be (6 + root number two) divided by two.

In the second question, the difference between the two sides of the triangle is less than that of the third side, and the P point is on the AF line.

In the third question, the length of pf1 and pf2 is the same as that of the first question, that is, pf1=(x+4)e, pf2=(4-x)e (f1 is the left focus, f2 is the right focus) (x+4) multiplied by (4-x) divided by 4 maximum4 minimum 3

Please explain the topic clearly by yourself, and you are too lazy to type, won't it make others think that you are not sincere?