High school math conic curve, guidance for academic masters

1. c a=1 2 a=2c b=root number c s f2ab=1 2 bc=root number 3 2 c=1 a=2 b = root number 3

2. (1) The parabolic equation is y 2=4x, and the linear equation is x=my+1, and the simultaneous linear equation and the parabolic equation can be obtained.

y^2-4my-4=0 y1+y2=4m

f1p=(x1+1,y1) f1q=(x2+1,y2)

According to the inscription, we can know that so x2=1 x1

x1+1=rx2+r x1+1=r x1+r gives x1=r x2=1 r

2) According to the meaning of the question, x1=r, so y1=2, r2=1, r2=2, r, r

pq|=(r-1 r) 2+(2 r-2 r) 2=r 2+1 r 2-2+4r+4 r-8

r is above (2,3).

So when r=3, the function obtains a maximum value of 112 9

r=2 to obtain a minimum value of 9 2

So |pq|The value ranges from (9 2 to 112 9).

Since I can't type the Greek alphabet here, I used a capital r instead.

Also because the picture is a little blurry, I estimate that F1, F2 is not wrong, if there is a mistake, you can ask, I can answer for you as needed, I hope this is the correct answer, I wish you progress in learning.

It is known that the point m(-2,0) and the point p of n(2,0) satisfy the condition |pm|-|pn|= 2 times the root number 2, and the trajectory of the moving point p is the equation 2 for finding cIf a, b are different points on c, and o is the coordinate origin, find the vector oa multiplied by the minimum value of the ob vector.

Labels: oaob, origin, vector.

Answer (1) Let the p coordinates (x, y).

pm|-|pn|= 2 root number 2

Root number [(x+2) 2+y 2] - root number [(x-2) 2+y 2] = 2 root number 2

To simplify: w is a hyperbola.

According to the definition: c = 2, 2a = 2 root number 2, c 2 = a 2 + b 2

b^2=4-2=2

Then the w equation is: x 2 2-y 2 2 = 1(x<0)

2) When the slope of the line ab does not exist, let the equation of the line ab be x x0, where a(x0,),b(x0,

When the slope of the line ab exists, let the equation of the line ab be y kx b, and substitute it into the hyperbolic equation.

Medium, get: (1 k2)x2 2kbx b2 2 0

According to the meaning of the problem, we can see that there are two unequal positive roots of equation 1°, let a(x1,y1),b(x2,y2), then.

Solution|k|>1, again.

x1x2＋y1y2＝x1x2＋（kx1＋b）（kx2＋b）＝（1＋k2）x1x2＋kb（x1＋x2）＋b2＝

To sum up. The minimum value is 2

Dahl. Rate: Solution: 1Because |pm|-|pn|= 2 times the root number 2, the point m(-2,0),n(2,0), then mn=4>2 times the root number 2, so the trajectory is c is the right branch of the hyperbola with the point m(-2,0), n(2,0) as the left and right focus.

then c = 2, a = root number 2, b = root number 2So the equation for c is x 2 2-y 2 2 = let a, b coordinates are (x1, y1), (x2, y2), then the vector oa multiplied by ob vector = x1*y1 + x2*y2. According to geometric analysis, when the vector OA is perpendicular to the ob vector, the value of the vector OA multiplied by the ob vector is the smallest, which is 0

Anonymity. Rate: 11 stick on the upper curve. c = the root equation is: x -y = 2 (x > 0).

Let the equation for ab be . x=ky+t,a(

The simultaneous equation is solved to obtain (k -1) y + 2kty + (t -2) = 0

y1y2=(t^-2)/(k^-2).

y1+y2=-(2kt) (k -1) vector.

When <> square, the numerator squared gets 121, and the denominator squared gets 4a, why is the denominator still 2 here?

This solution is also incorrect, you take x2=5, y2=6 back to the equation and do the math, it can't satisfy the equation of the circle.

Generally, the chord length is calculated by the distance and radius from the center of the circle to the straight line, and there is no need to calculate the specific two points, as shown in the figure below.

x = (2 - 14) 2 or x = (2 + 14) 2 when x = (2- 14) 2 when y = (4 - 14) 2 and when x = (2 + 14) 2y = (4 + 14) 2

ab = 14) +14) ]56 = 2 7 Another solution: the straight line is; x-y+1=0

x²+y²-4x-2y-4=0

x-2)²+y-1)²=9

The coordinates of the center of the circle (2,1), the radius r=3, let the distance from the center of the circle to the straight line be d, d= 2-1+1 2= 2, ab =2 (r -d )=2 7

Method 1: Make use of algebra.

Simultaneous y=kx+2 and x-y=6 yields (k-1)x +4kx+10=0

The straight line intersects the right branch of the hyperbola at two different points, and the equation for x has two unequal positive roots.

So there is δ=(4k) -40(k -1)>0

and x1+x2=4k (1-k)>0, x1x2=10 (k -1)>0

Respectively solved - 15 31

So- 15 3 Method 2: Use the geometric method.

The straight line y=kx+2 passes the fixed point (0,1), draws the straight line and the hyperbola, and the crossing point (0,1) is a parallel straight line with the asymptotic line y= x of the hyperbola.

It is observed that the line tangent to the right branch of the line (0,1) and the line parallel to the line y=-x and the parallel line y=-x+1 have two different intersections with the right branch of the hyperbola.

Find the slope of the tangent with the right branch as -15 3, so the range of k is -15 3

y=kx+2

x²-y²=6

k=0, it is true.

k≠0: simultaneous equation.

1-k²)x²-4kx-10=0

Intersection at two different points, i.e., =(16k)+40(1-k)>0, the solution is now obtained: - 15 3

af1+af2 = bf1+bf2 = 2a = 10 (ellipse definition), so the perimeter of abf2 is c = 20

The circumference of the inscribed circle is

So the radius of the inscribed circle is.

So the area of abf2 s = * r * c = 5 (the area obtained by dividing the triangle).

and the area of abf2 s = * f1f2| *y1-y2| = * 6 * y1-y2| =3|y1-y2|

So |y1-y2|=5 out of 3 choose d

The relationship between the hyperbolic focal point and the real axis and the imaginary axis is: c 2 = a 2 + b 2, from which we can find the beam without bond coordinates of the left Jiaocha dry point a, and the coordinates of the f point are (-c, y), which is substituted into the hyperbolic equation to find the value of y. The value of y is the distance of af.