A problem of math ellipses in the second year of high school, help solve it

Ah, let's find the elliptic equation first, yes, 9/x square + y square =1 and then let the equation of the straight line l be y=kx+m, because (-2,0), and k=1, bring the point into the linear equation. Then we can find out that l is y=x+2

Then the equations of the simultaneous ellipses and lines. A comprehensive equation can be obtained. From Veda's theorem we can find x1 + x2, x1 times x2

Then ab uses the chord length formula, ab = under the root number [(1+ksquare)[(x1+x2)square-4x1x2] dangdangdang - the first question is like this

Then the second question, let m(x,y).

x = x1 + x2 of 2

y = y1 + y2

y1 and y2 can be expressed as x1 and x2 respectively according to the linear equation. So y can also be represented by x1 and x2.

Then m is on the straight line l, just bring in the equation

a=6*2=3 c=2 2 then b=a2-c2=1 then the elliptic equation is x2 9+y2=1 the straight line l passes through the point (-2,0) and the slope is 1, then the equation is y=x-2 generation x2 9+y2=1 to get x2 9+(x-2)2=1 x1x2=?ab= k2+1 is being used

x1+x2)2-4x1x2=?

Let a(x1,x1-2)b(x2,x2-2) then xm=x1+x2 2 ym=x1-2+x2-2 2 x1+x2=? If we know, then we can know the trajectory equation of the point m in xm,ym for y=x-2.

where x1+x2= x1x2= is derived using the Weida theorem.

(2 Let p(x0,y0), a(3,0), m(9 2,ym) pass the point p to make pb perpendicular to af, let the intersection of the right alignment and the x-axis be n, then pb:mn=fb:fn

i.e. y0 ym=(x0+2) (9 2+2), i.e. ym = (13y0 2) (x0+2).

k1=y0/（x0-3），k2=ym/（9/2-3）k1·k2=y0/（x0-3）*ym/（9/2-3）=2y0ym/[3（x0-3）]

13y0*y0/[3（x0-3）（x0+2）]x0^2/9+y0^2/5=1，y0^2=5/9（9-x0^2）k1·k2=（65/27）*（9-x0^2）/[x0-3）（x0+2）]

(65 27)*(x0+3) (x0+2)=-(65 27)*[1+1 (x0+2)]fm intersects ellipse c at p,-21 5,1+1 (x0+2)>6 5-(65 27)*[1+1 (x0+2)]<26 9k1·k2<-26 9

Let p(x1,y1)(-2, so y1 (x1+2)=y2 (13 2), so y2=13y1 (2(x1+2)).

So m(9 2,13y1 (2(x1+2)) because k1=y1 (x1-3) k2=13y1 (3(x1+2)) so k1*k2=13y1 2 2(x1+2)(x1-3) because p is on the ellipse c.

So x1 2 9+y1 2 5=1

So y1 2=-5 9(x1 2-9).

So k1*k2=-65 27*(x1+3) (x1+2)=-65 27*(1+1 (x1+2)).

Because -2, k1*k2<-26 9

I'm assuming p(3cost, root 5sint) -90 degrees less than t less than 90 degrees, and t is the parameter. The final answer is (-26 roots, 5 45, -13 roots, 5 18), I don't know if it's right or not.

The area of the whole rectangle is the largest, and it can be seen from the elliptic symmetry that only the first quadrant rectangle s is maximum, s=xy

The elliptic equation 1 = x 2 100 + y 2 64 is greater than or equal to (mean inequality) under x 2y 2 6400 under 2 times the root number, and the xy is less than or equal to 1600, if and only if x 2 100 = y 2 64 takes the equal sign,,, and the elliptic equation is combined to obtain x = 5 times the root number 2, y = 4 times the root number 2

At this time, the perimeter c = 4 (x + y) = 36 times the root number 2

Let p(5cost,4sint).

Left and right focal points of the ellipse: f1 (-3,0), f2 (3,0).

pf1=(-3-5cost,-4sint)pf2=(3-5cost,-4sint)

pf1*pf2=25cos²t-9+16sin²t=9cos²t+7≥7

(pf1·pf2)min=7

If it is known that a=5, b=4, then c=3

Let p(x,y), then there is the vector f1p=(x+3,y) and the vector f2p=(x-3,y).

and (x 2+6x+9+y 2)+ x 2-6x+9+y 2)=10

Next, find the minimum value of x 2-9+y 2.

x=0, y=4, and the minimum value of vector pf1*vector pf2 is 7

Formation of the general formula x 2 4+y 2 2=1 a= c= 2 left focus- 2,0 inclination angle 60 slope is 3 substituting the left focal point to get the linear equation y= 3 x+ 6 substituting the elliptic equation to get 7x 2+12 2 x+8=0 Vedic theorem, write the relationship between the two roots, substitute the chord length formula (x1+x2) 2-4x1x2>(1+k 2). x1x2=8 7 x1+x2=-12 2 7 k is 3, I get 16 7 The second question finds the absolute value of the solution of the first question y and multiplies the focal length by one-half The third question is the longest at its perpendicular bisector. Since the two straight lines are perpendicular to each other, the slope product is -1, so the straight line is y=- 3- 6, substitute into the elliptic equation to find the intersection point, um.

A value should be rounded.

Solution: From the problem a=6 , b=2 5, c=4

a(-6,0) b(6,0) f(4,0) Let p(x,y) where y >0

Vector (pa·pb) = 0.

6-x,-y)·(4-x,-y) =0

i.e. x 2+2x+y 2-24=0 .1)

Synact x 2 36+y 2 20=1....2)

It can be solved x=3 2 y=5 3 2>0

i.e. p(3 2,5 3 2).

Let the linear ap equation be k=(5 3 2) (3 2+6)= 3 3, 3x-3y+6 3=0

Let m(x,0), |mb|=6-x

The straight-line distance from M to AP is d1=|√3x+6√3|/√12=6-x

The solution is x=2

The point q (6cos, 2 5sin) on the ellipse is to m(2, 0) at a distance d

d 2= (6cos -2) 2+(2 5sin) 2 where [0,2].

16cos²θ-24cosθ+24

16t -24t+24=(4t-3) +15>0 where t [-1,1].

It can be seen that when t=3 4, there is min(d2)=15

min(d)=√15

The answer process is relatively long, see the information:

From a=2c,a 2 c=4,a=2,c=1,b 2=3, the elliptic equation 3x 2+4y 2-12=0, let m(s,t), pass m to make me perpendicular x-axis to e, m to do md parallel pb and tremble to intersect x-axis to d, alignment and x-axis to q, de:bq=me:pq=ae:

aq，de=1/3ae=1/3（s+2），eb=2-s。Only the hole Zheng needs to prove that the angle NBM is an obtuse angle, that is, the angle DBB = angle MBP is an acute angle, that is, only MD 2+MB 2>DB 2, that is, DE 2+EB 2+2*ME 2>DE2+EB 2+2*DE*EB, that is, ME 2>DE*EB, this question can be proven.

m(s,t) is on an ellipse and t<>0, de*eb=1 3(4-s2)=1 9(12-3s2)=4 9t2