A high school math problem, ellipse. Ask for advice 20

The line y=2x+8 is moved parallel to the tangent position in the direction of the ellipse, and the distance from the tangent point to the line y=2x+8 is the shortest distance. The tangent point (x0, y0) is the midpoint of the chord when the chord length is 0, and the spread method can be obtained by multiplying (y0 x0) by the slope of the straight line 2=(-b 2 a 2), so y0 x0=-1 generation of the ellipse 2x 2 + y 2=2 equation (x0, y0) = is obtained

6 3, 6 3) to the straight line 2x+8-y=0 is (8 5- 30) 5

It is also possible to set the tangent to y=2x+t, and the discriminant formula =0 to obtain t= 6 (or t=- 6 is rounded up to the farthest), then the distance between the parallel lines of y=2x+t and y=2x+8 (8, 5- 30), 5 is the sought.

Solution: The shortest distance from a point on an ellipse 2x 2+y 2=2 to a straight line y=2x+8 can be reduced to the distance between lines parallel to the line y=2x+8 and having a unique intersection with the ellipse.

Let the line y=2x+t have a unique intersection with the ellipse 2x 2+y 2=2, then there is 2x 2+(2x+t) 2=2

Simplified: 6*x 2+4*t*x+(t 2-2)=0 We have: 16t 2-4*6*(t 2-2)=0 solution: t= 6 or t=- 6

Let the inclination angle of the straight line y=2x+8 be , then the minimum distance between the two straight lines y=2x+ 6, y=2x- 6 and the straight line y=2x+8 is .

8-√6)*cosα=(8-√6)*/√(1+(tanα)^2)=(8-√6)/√5=√5*(8-√6)/5。

The point with the shortest distance from the ellipse to the line y=2x+8 is the tangent point between the parallel line of the line y=2x+8 and the ellipse.

The elliptic equation 2x 2 + y 2 = 2 is derived: 4x+2yy'=0 => y'=-2x/y

by y'=-2x y=2 gets: x=-y, and substituting the elliptic equation gets: 3x =2 =>x=-y= 6 3

Obviously, the point in the second quadrant (- 6 3, 6 3) is the closest point to the straight line, and the point in the fourth quadrant ( 6 3, - 6 3) is the farthest point to the straight line.

Make use of the distance formula from a point to a line:

Minimum distance dmin=|(-2√6/3 - 6/3 +8|/√5=(8√5-√30)/5

Let x=cosa

Then the range of y 2=2-2(cosx) 2=2(sina) 2sina is symmetrical with respect to the origin, so you might as well make y=2sina so the point on the ellipse is the straight-line distance =|2cosa-√2sina+8|/√(2^2+1^2)

2cosa-√2sina=-√2sina+2cosa=-√(4+2)*sin(a+z)=-√6sin(a+z)

where tanz = 2 2 = 2

So sin(a+z)=1.

2cosa-√2sina+8|Minimum = 8- 6, so minimum distance = (8- 6) 5=(8 5- 30) 5

Let p(x,y), pf1 pf2 = the distance from p to the left alignment x the distance from the right alignment x the square of the eccentricity

x is between 2 and 2, desirable, etc.

When x = 0, the maximum is 4, and when x plus or minus 2, the minimum is 3, and I hope it can help you.

The minimum product of PF1 and PF2 is 3 (in fact, the area of the inscribed triangle with the radius of root three), F1F2P

On the same line, the product is maximum, four plus two, three, a=2b=1

(1) A 2 = 2 b 2 = 1 c = 1 Let the l equation.

bai is y=- root number.

du2*x+1 a(x1,y1) b(x2,y2) p(x0,y0)

Substituting the l equation into the zic equation and reasoning: 4x dao2-2 root number 2x-1=0x1+x2=2 root number 2 y1+y2=-root number 2 (x1+x2)+2=-3

OA+OB+OP=(x1+x2+x0,Y1+Y2+Y0)=(2 roots, 2+X0,-3+Y0)=(0,0).

x0=-2 root number 2, back to y0=3, that is, p(-2 root number 2,3) can verify that the coordinates of p point are satisfied.

Answer: l equation. 2) q (2 root number 2,-3).

Left focal point f1(-1,0),b(0,1), so ab:y=x+1, coupled with the elliptic equation, gives a(-4 3,-1 3), the area of abf2 = 1 2*2(1+1 3)=4 3

Ellipse: x 4 + y 3 = 1

The use of vectors is actually the use of the fixed-point coordinate formula, this formula is said to be no longer learned in the current textbook, but appears in the after-class exercises, I will give you a derivation, you can use it directly during the exam.

Let p1(x1,y1),p2(x2,y2),p(x,y), if p1p = pp2 (1), then it is called the ratio of the point p to the directed line segment pp1, p is called the fixed score point.

p1p→=(x-x1,y-y1),pp2=(x2-x,y2-y)

Because p1p = pp2

So x-x1= (x2-x), y-y1= (y2-y).

x=(x1+ x2) (1+)y=(y1+ y2) (1+) This is the coordinate formula for the fixed score point p.

As an aside, when p is the midpoint =1, substitute it into the formula to get p((x1+x2) 2,(y1+y2) 2), this is called the midpoint coordinate formula, you should have learned it, right?

Returning to this problem, the ratio of a(2,0) to the directed line segment nm =12 7, let n(x1,y1),m(x2,y2), according to the coordinate formula of the fixed score point, 2=(x1+12 7*x2) (1+12 7), 0=(y1+12 7*y2) (1+12 7).

x2=19 6-7x1 12,y2=-7y1 12

Substituting the equation of the circle, we get 7x1 +7y1 -4x1=20

And because n is on an ellipse, it satisfies 3x1 +4y1 =12

The solution is x1 = 2 7 or 2 (round).

So y1 = 12 7 or -12 7

When n(2 7,12 7), kan=(12 7-0) (2 7-2)=-1, so l:x=-y+2

When n(2 7,12 7), kan=(12 7-0) (2 7-2)=-1, so l:x=-y+2

Later on, the calculation will be simplified.

Three unknowns, three equations to solve, and then you can find it yourself!

(1) Set p(x,y) a(xa,2 5xa)b (xb,2 5 5xb).

Because ab = root number 20

So (xa-xb) square + 2 root number 5 5 (xa + xb) square = 20 substitution x = xa + xb y 2 root number 5 5 = xa-xb to get it.

2 5 5 x squared + 5 4 y squared = 20 (the trajectory is an ellipse) (2) The known trajectory equation Let xm xn solve it.

You should be able to do this kind of question in the first question, and I won't do it in the second question! Hey...

You might as well let m(a 2 c,y1),n(a 2 c,y2), from f2(c,0) and f2m and f2n give f2m 2+f2n 2=mn 2, i.e., (a 2 c-c) 2+y1 2+(a 2 c-c) 2+y2 2=(y1-y2) 2, and simplify b 4 c 2=-y1*y2.

and c(a 2 c,(y1+y2) 2), r=|y1-y2|2, so, by.

oc^2-r^2

(a^2/c)^2+(y1+y2)^2/4]-(y1-y2)^2/4

a^4/c^2+(y1^2+y2^2+2y1*y2)/4-(y1^2-2y1*y2+y2^2)/4

a^4/c^2+y1*y2

a^4/c^2-b^4/c^2

a^2+b^2

0, oc>r can be obtained, that is, the origin o is outside the circle c.

In fact, the circle c is constant over f2, while the line x=a2 c is the right alignment of the ellipse, o is to the left of f2, and therefore o is outside the circle c).

The center of the circle must be on the right alignment, F2M is perpendicular to F2N, and F2 is on the circle, and it is obvious that any point on the alignment is closer to F2 than to Point O, so it is outside the circle.

Since vectors are not easy to write and look at, you can see the big picture as long as you double-click on the diagram, thank you

Take advantage of one of the geometric properties of the ellipse: the product of the slopes of any point on the ellipse and the two ends of the major axis is b 2 a 2

That is, in this problem, kpm·kpn 1 4, the slope of the other pn can be solved in the range of 1 2, -1 8

The value range of the slope of the straight line pm is ?

The product of the slope of the line connecting any point on the ellipse and the two ends of the major axis is b 2 a 2

In this problem, kpm·kpn 1 4, the slope of the other pn can be answered in the range of 1 2, -1 8