Sophomore math, conic curves, to detailed steps.

1. Solution: Defined by the ellipse: absolute value pf1 + absolute value pf2 = 2a from the question: absolute value pf1 = 4 3 , absolute value pf2 = 14 3 So, 2a = 4 3 + 14 3 = 6

So, a=3

And because, pf1 f1f2 , f1f2=2c, pf1f2=90°

In RT pf1f2, the Pythagorean theorem is used: f1f2 = pf2 -pf1 = (14 3) -4 3) = 20

2c)²=20 c²=5

So, b = a -c = 4

Therefore, the standard equation for the ellipse is x9+y4=12, let the coordinates of the two intersections of the line l and the ellipse a(x1,y1) b(x2, y2), x1 9+y1 4=1---1).

x2²/9+y2²/4=1---2)

1)-(2) gets: (x1 -x2 ) 9+(y1 -y2 ) 4=0y1+y2)(y1-y2) (x1+x2)(x1-x2)=-4 9---3).

Because, the points a(x1,y1) and b(x2, y2) are symmetrical with respect to the center of the points m(-2, 1).

Therefore, (x1+x2) 2=-2, (y1+y2) 2=1 and because of the slope of the straight line k=(y1-y2) (x1-x2), from (3) we get: k=8 9

Therefore, the equation for the straight line is: y-1=(8 9)*(x+2), i.e., y=(8 9)*x+25 9 or 8x-9y+25=0<>

Conic curves include ellipses, hyperbolas, and parabolas, so what are the detailed steps?

In-plane with two fixed points f, f'The sum of distances is equal to the constant 2a(2a>|ff'|The trajectory of the moving point p is called an ellipse.

i.e.: PF + PF'│=2a

Two of them are fixed points f and f'It is called the focal point of the ellipse, and the distance between the two focal points ff'The focal length is called an ellipse.

The set of points on a plane where the ratio of the distance to the fixed point f to the distance to the fixed line is constant (the fixed point f is not on the fixed line, the constant is a positive number less than 1).

where the fixed point f is the focal point of the ellipse, and the fixed line is called the alignment of the ellipse (the equation for the fixed line is x=a2 c).

Alignment: ellipse and hyperbola: x=(a2)c

Parabola: x=p2 (take y2=2px as an example) focal radius:

Ellipses and hyperbolas: a ex (e is the eccentricity. x is the abscissa of the point, less than 0 takes the plus sign, and greater than 0 takes the minus sign).

Parabola: p 2 + x (take y 2 = 2 px as an example) above the ellipse and hyperbola with the focus on the x-axis as an example. Chord length formula:

Let the slope of the straight line where the string is located be k, then the chord length = root number [(1+k 2)*(x1-x2) 2] = root number [(1+k 2)*(x1+x2) 2-4*x1*x2)] The equation of the straight line is connected with the equation of the conic curve, and the unary quadratic equation about x is obtained by eliminating y, x1 and x2 are the two roots of the equation, and x1+x2 and x1*x2 can be known by using Veda's theorem, and then the chord length can be obtained by substituting the formula. Parabolic diameter = 2p parabolic focal chord length = x1 + x2 + p Combining the equation of the focal chord with the equation of the conic curve, subtracting y is to obtain a quadratic equation about x, and x1 and x2 are the two roots of the equation.

In-plane with two fixed points f, f'The sum of distances is equal to the constant 2a(2a>|ff'|The trajectory of the moving point p is called an ellipse.

i.e.: PF + PF'│=2a

Among them, the split faction has two fixed points f and f'It is called the focal point of the ellipse, and the distance between the two focal points ff'The focal length is called an ellipse.

The set of points on a plane where the ratio of the distance to the fixed point f to the distance to the fixed line is constant (the fixed point f is not on the fixed line, the constant is a positive number less than 1).

where the fixed point f is the focal point of the ellipse, and the fixed line is called the alignment of the ellipse (the equation for the fixed line is x=a 2 c).

By the title 2|mf|=|af|+|bf|=|ab|The straight ab passes through the focal point (2,0).

Let the ab equation be ky=x-2

By the inscription y = 8x, ky = x-2

Eliminate x, y -8ky-16 = 0, y1 + y1 = 8k, y1y2 = -16

x1+x2=k(y1+y2)+4=8k²+4|ab|=8(1+k²)

I don't want to solve the rest, you can figure it out yourself.

With 2|mf|=|ab|, you can get the relationship between k and x0, y0 in the midpoint coordinates, then the perpendicular bisector equation can be expressed, and then n point abscissa can be found n point to represent k can.

denotes n

Looking back now, it was a real headache ......

== Generally, we can set the coordinates of two points, and we can generally set a straight line, and then generally, we can eliminate y to get the 2nd order inequality of x, and then Weida's theorem x1+x2=. x1x2=..Look at the question again, what do you want to ask for = = is generally to find the equation of x y with x with y - and finally bring it down can be brought with x1 + x2 or x1x2 If it doesn't work, this question is directly lost = =

Hello classmates, I am Guo Limei, a teacher from New Oriental Youneng Learning Center.

Ellipses and hyperbolas are similar in conic curves, and parabolas are treated separately.

First of all, for ellipses and hyperbolas, it is necessary to grasp the definition, eccentricity, and the relationship between the focal point and the standard equation (a, b, c). As for finding equations, it is generally necessary to use eccentricity, coordinates of points, asymptote and other conditions to list the equations of the relationship between a, b, and c, and solve the equations.

Secondly, for the comprehensive problem of conic curves, it is necessary to learn the traditional method (Vedder's theorem) and the spread method, use the coordinates of the two intersection points to represent the conditions in the problem, practice some basic problems first, and master the methods and ideas of problem solving. However, the amount of calculation is generally large, so you must first do a few questions carefully and summarize the method.

Finally, you can find a teacher, practice more types of questions, and let the teacher explain the method to you, the conic curve is generally in the penultimate question in the college entrance examination, which has a certain degree of difficulty, but the method and law are also followable, often integrated to vectors, functions, etc., in the last review stage, more practice must be helpful.

Good luck.

Let x=cosa of 2/2 of the root number, y=sina, then the point to the straight-line distance formula d=i root number 3x-y-4i 2, and substitute x,y into d=(root number 10+8) 4 when sina is taken as the minimum value

1) It is known that the circle is a circle with a radius of a(-3,0) r=10, and the equation of the moving circle is (x-a) +y-b) =r, so the center of the circle is the moving point, and it is set to c(a,b).

Then by the tangent condition: ac=(a+3) +b =(10-r) by b(3,0) on the moving circle: (3-a) +b =r

(a+3) -a-3) =(10-r) -r gives 12a=100-20r

From this, we get r=5-(3 5)a, and substituting (a-3) +b =[5-(3 5)a] to get a 25+b 16=1

This shows that the trajectory of the center of the moving circle c(a,b) is elliptical.

2) If 01, then there is no point where the absolute value of the difference between the distance from f1 and f2 is the fixed value a, and the trajectory of such a moving point p does not exist.