Urgent!! High School Ellipse Questions! Ellipse problem! Urgent!!

In mathematics, an ellipse is a trajectory where the sum of the distances from two fixed points on a plane is constant. These two fixed points are called focal points. x 2 12+y 2 3=1,, the focal point is (3,0)(-3,0)f1f2=6f1f2 is the origin o The line mo is the median line of the triangle pf1f2 mo f1f2 (y-axis perpendicular to x-axis) pf2 f1f2 (median line parallel to the bottom line) p x coordinate is x coordinate of f2 in ellipse x 12 + y 3 = 1 a 2 = 12 (a=2 3) b 2 = 3 (b= 3) c 2 = a 2 -b 2 = 9 c = 3 The coordinates of f1 are (-3,0) The coordinates of f2 are (3,0), and by bringing x=3 into the elliptic equation, the absolute value of the coordinates of p point y is :

y| = √[3（1-x^2/12)]=√3/2∴|pf2|= 3 2 is defined by the 2nd of the ellipse, there is a |（|pf1|+|pf2|）|=2a∴|pf1|=2a-|pf2|=2*2√3-√3/2=7√3/2∴|pf1|/|pf2|=7:1 is 7 times.

Equation with ellipse: x 2 + (y 2) 3 = 1

It can be seen that the focal point is located on the y-axis, the coordinates f1(0,-2), (0, 2) let the intersection points a(x1,y1) and b(x2,y2) of the line and the ellipse be obtained by the midpoint coordinate formula

k+1)(x1+x2)+2b=4

Then substitute b 2=[(k 2+3) 2] k 2+9.

You can solve the relationship between K and Huyou x1x2.

First of all, AP and BP intersect x=3 at mn

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The standard equation for the ellipse is x 20 + y 16 =1a=2 5,b = 4,c = (20-16)=2 so the point b coordinate (0,4), the right focal point f coordinate (2,0) sets the point m coordinate (x1,y1), and the point n sits on the year to guess the mark (x2,y2) Since the point f is the center of gravity of the hall year BMN, so 2=(x1+x2+0) 3,0=(y1+y2+4) 3 [center of gravity coordinate formula].

The solution yields x1+x2=6, y1+y2=-4

So the MN midpoint coordinates are pretending to be finches' eyes (3,-2).

The points m, n are both on the ellipse, satisfying 4x1 +5y1 =80, 4x2 +5y2 =80, and the two equations are subtracted to obtain:

4(x1-x2)(x1+x2)+5(y1-y2)(y1+y2)=0, so the slope of the straight line l = (y1-y2) (x1-x2) = (4 5) (x1+x2) (y1+y2) = (4 5) (6 -4) = 6 5

The midpoint of Mn (3,-2) is on the line L, and the equation L is obtained from the point oblique

y+2=(6/5)(x-3)

That is: 6x-5y-28=0

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Solution: easy to know, focus f1(-c,0)Straight line l:

y=k(x+c).===> point c(0,kc), and then b(-c 2,kc 2)And because the point b is on the ellipse, [c (4a )]k c (4b )]=1

=>a²≥2c².===>0＜e≤(√2)/2.

Solution: Op=1 2pq=1 2pb and op+pb=b to obtain op=1 3b

From the meaning of the title, we can get op=c=a2-b 2, i.e., 1, 3b=a2-b2 (1).

The intersection point of the tangent line and the circumscribed circle in the problem is g, and the intercept coordinate on the x-axis is h(-(3 2) 4,0).

In RT BOH, ob*oh=bh*og ob=b,oh=3 2 4,bh=(b 2+9 8) (1 2), og=c=1 3b

Bring in the data and bury the traces to obtain b*3 2 4=(b 2+9 8) (1 2)*1 3b (2).

Solution, crack and b=3

Substitute liquid argument (1), get a2 = 10

The elliptic equation is x 2 10 + y 2 9 = 1

(This number is cumbersome to calculate.)

This problem does not seem to be difficult to know the slope of the straight line, but also know that it is tangent to the garden, the equation of the straight line is very good, from the origin to the tangent point, as shown in the figure, first find the equation of the straight line, the equation of the garden knows, indicating that the radius is with the sign 3, that is to say, the distance from the origin to the straight line is the root number 3, because the angle is 45°, so you can find the equation of the straight line,

Solution: (1) Let the straight line l:y=x+n be substituted into a circle, and 2x 2+2nx+n 2-3=0

Tangent =4n 2-8n 2+24=0 n=-root number 6 (root number 6 rounded) straight line l; y=x - root number 6

2) Straight line substitution ellipse: 7x 2-8 root number 6x+12=0 x=(4 root number 6+-2 root number 3) 7, y=(-3 root number 6+-2 root number 3) 7

a((4 6-2 3) 7, (3-2 3) 7)b((4 6+2 3) 7)

The focal point f(1,0) is obtained by (1) q(root number 6 2, -root number 6 2) |af|=(96-48 root number: 2+12) 49-(8 root number, 6-4 root number, 3) 7 + 1 + (54 + 36 root number, 2 + 12) 49 =

Let a(x,y)(x>0), then |af|²=(x-1)²+y²=x²-2x+1+3-(3/4)x²=4-2x+(1/4)x²=(2-x/2)²

So, |af|=2-x/2。

Again, |aq|²=|ao|²-oq|²=x²+y²-3=x²+[3-(3/4)x²]-3=x²/4

So, |aq|=x/2

So, |af|+|aq|=2。

In the same way, |bf|+|bq|=2。

So, |af|+|aq|=|bf|+|bq|。

Let q(m,n), then m +n = 3

By oq straight line l, then the slope of l is -m n

The equation is y-n = -m n · x-m).

i.e. y=n+m n-mx n

y=3/n-mx/n

Substituting elliptic equations.

3x²+4(3/n-mx/n)²=12

4m²/n²+3)x² -24m/n²·x +36/n²-12=0

Coordinates of the right focal point f (1,0).

af| = √[x1-1)²+y1²] =√[(x1-1)²+3(1-x1²/4)] = √(x1²/4-2x1+4) = √[x1²-8x1+16)/4]

[(x1-4)²/4] = (x1-4)/2

aq| = √[x1-m)²+y1-n)²]= √[x1-m)²+m/n · x1-m))²= √[x1-m)²(1+m²/n²)]

[x1-m)²(3/n²)]= √3·(x1-m)/n

af| +aq |= (x1-4)/2 + 3·(x1-m)/n = (1/2+√3/n)x1 - 2+√3m/n)

In the same way, |bf|+|bq| = (1/2+√3/n)x2 - 2+√3m/n)

and x1+x2 = (24m n) (4m n +3) = 24m (4m +3n ) =24m (m +9).

1/2+√3/n)x1 - 2+√3m/n)

1/2+√3/n)x1 - 2+√3m/n)

Here's an idea. With f as the pole, a polar coordinate system is built, and two angles are set to represent three points, all of which are in a straight line, and the polar coordinates of the ellipse and the line are comprehensively solved. There is a formula for the tangent of the circle, x0x y0y-r2 0, and then translate it to point f.

Same as the second year of high school, math is terrible, and it seems like a stupid way to find the coordinates of a, b, q, and f.

Analytic hair is to write out all the analytic coordinates.

Under the heavy reward, there will be brave men!

Answer: Compared with the standard form of an ellipse, the curve represented by the equation (3m+7)x2+(3m+4)y2=5m+12 is an ellipse, then 3m+7≠3m+4 (obviously true)(5m+12) (3m+7)>0 is m<-12 5 or m>-7 3

5m+12) (3m+4)>0 i.e. m<-12 5 or m>-4 3

In summary, the value range of m is m<-12 5 or m>-4 3